Download 1. Blood cholesterol levels are generally expressed as milligrams of

1. Blood cholesterol levels are generally expressed as milligrams of cholesterol per deciliter of
blood. What is the approximate mass percent of cholesterol in a blood sample having a
cholesterol level of 176? Why can you not give a more precise answer? Chemistry
ANSWER: mass % = mass of cholestrol/100 gm blood
we take blood density = 1.0
100 ml blood means 100 grams
for 100 ml i.e 1 deciliter (= 100 gm) blood ,cholestrol mass = 176 mg =0.176 gm
mass % = 100x0.176/100 = 0.176
density of blood is not known precisely , so answer cannot be given precisely
2. The temperature inside a pressure cooker is 1150C. Calculate the vapor pressure of the
water inside the pressure cooker. (Hvap= 40.7 kJ/mol ) The normal boiling point of water
is 1000C. (Hint normal is a chemistry term.)
ANSWER: Write out all the variables.
p1 = pressure at normal boiling point = 1.0 atm
p2 = 3.5 atm
ΔHvap = enthalpy of vaporization = 40,790 JOULES/mole
R = constant = 8.31 J / mole K
T2 = ?
T1 = normal boiling point of H2O = 373 K
ln (1.0 / 3.5) = (40,790 / 8.31)(1/T2 - 1/373)
-1.25 = 4909 (1/T2 - 1/373)
-1.25 = 4909/T2 - 13.2)
11.9 = 4909/T2
4909/11.9 = T2 = 413 K = 140 C
3. The
decomposition of formic acid follows first-order kinetics: HCO2H(g) ? CO2(g) + H2(g)
The half-life for the reaction at 550 �C is 24 seconds. How many seconds does it take for the
formic acid
concentration to decrease by 87.5% (12.5 % remains)?
ANSWER: there are several equivalent equations for half-life
but
this is the one you will see most often
N/No = e^(-0.693*T/Tk)
(math nerds love this one because "e" is so convenient to use in the calculus)
N is present amount and No is original amount
in this problem N/No = 0.875
and Tk = 24 s
so
0.875 = e^(-0.693*T/24)
ln(0.875) = -0.693 T / 24 (ln is natural log)
T = 4.62 s
the other common equations are:
N/No = 10^(-0.301 T/Tk)
or
N/No = (1/2)^(T/Tk)
4. I have been looking at the practice questions in my chemistry book and seem to not get the
same answer as they have. The question is: How many oxygen atoms are in 0.25 mol Ca(NO3)2.
The answer is 9.0x10 to the 23rd power and I get 1.44x10 to the 25 power. What am I doing
wrong?
ANSWER: this
is exact way to solve
Ca(NO3)2
there are 3 x 2 = 6 atoms of oxigen in 1 molecule
0.25 x 6 = 1.5 moles of oxigen
Na2CO3
there are 3 atoms of oxifen
1.50 x 3 = 4.50 moles of oxigen
5. Carminic acid, a naturally occurring red pigment extracted from the cochineal insect, contains only
carbon, hydrogen and oxygen. It was commonly used as a dye in the first half of the nineteenth century.
It is 53.66% C and 4.09% H by mass. Calculate the empirical formula for carminic acid.
ANSWER:. let 100 g of acid
53.66 g of Carbon
moles of carbon = 53.66/12 = 4.47
4.09 g of Hydrogen
moles of hydrogen = 4.09/1 = 4.09
42.25 g of oxygen
moles of oxygen = 42.25/16 = 2.64
ratio C:H:O = 4.47:4.09:2.64
rationalizing
C:H:O = 22:20:13
C22H20O13
formula
6. Cesium (cesium-137) has a half-life of 30 years. Suppose we have a 100 mg sample. a. find
the mass that remains after t years b. how much of the sample remains after 100 years? c. after
how long will only 1 mg remain? Please show steps not just answers using Calculus formulas not
chemistry formula
ANSWER: Plug this into the equation.
0.5 = e^(tk)
0.5 = e^(30k)
Take the natural logs of both sides.
ln(0.5) = ln e^(30k)
ln(0.5) = 30k(ln e)
ln(0.5) = 30k(1)
k = ln(0.5) / 30
k = -0.693 / 30
k = -0.023
A = Pe^(kt)
A = ending amount
P = starting amount
t = half-life time
k = decay constant
(a) Find the mass that remains after t years.
Plug in the known P and t values.
A = Pe^(kt)
A = (100)e^(-0.023 * t)
A = 100e^(-0.023t)
ANSWER: A = 100e^(-0.023t) mg
(b)
Plug in your known t value.
A = 100e^(-0.0231t)
A = 100e^(-0.0231 * 100)
A = 100e^(-2.31)
A = 100 * 0.125
A = 9.93 mg
c)
A = 100e^(-0.0231t)
1 = 100e^(-0.0231t)
t = 199 years.
7. I will post an easy principles of chemistry question based on conversations in 10 minutes from
now and i will have only 15 mins to answer it. Let me know right now if anyone will be
available at 12:01pm to be here and solve it.
ANSWER: Zn(s)+2HCl(aq)-->H2(g)+ZnCl2(aq)
21.15 g Zn / 65.38 grams/mol = 0.323 mol Zn
25.26 g HCl / 36.46 grams/mol = 0.623 mol HCl
Since we need 2 HCl per Zn then HCl is the limiting reagent, as we would need 0.646 mol HCl
to react completely
0.623/2 = 0.312 mol
0.312 mol * 136.315 grams/mol ZnCl2 = 42.53 grams
8. You need to produce 500.00ml of an aqueous solution in which the concentration of PB2+ is 10 mg/L.
Your source of PB2+ ions is lead (II) nitrate. How much lead (II) nitrate do you need to wieigh out to
produce this solution? This is analytical chemistry
ANSWER: Volume of solution = 500 ml = 500 /1000 = 0.5 L
Concentration of Pb+2 = 10 mg /L = 10 *10^-3 = 0.01 g/L
[ 1mg = 10^-3 g]
Concentration = grams / volume of solution in litres
0.01 = grams of pb+2 / 0.5 L
Grams of Pb +2 = 0.01 *0.5 = 0.005 g
Pb (NO3)2 --> pb+2 + 2NO31 mole pb(NO3)2 = 1 mole pb+2
Grams of Pb (NO3)2 =
( 1 mole Pb(NO3)2 / 1mole Pb) * (331.2 g Pb(No3)2 / 1 mole Pb(NO3)2 ) * (1 mole Pb /207.2
g Pb) * 0.005 g Pb(ii)
= 7.99 *10^-3 g = 7.99 mg Pb (NO3)2
We weigh out 7.99 mg g lead (II) nitrate to produce this solution
9.
ANSWER: Average and standard deviation is 93.503+\-0.014 %
standard error is 0.014/3^1/2= 0.008
95% confidence interval is 93.487 to 93.519%
10. Liquid nitrogen is an extremely cold liquid (-196 ?C) used in chemistry "magic" shows to
"shrink" balloons (and do other fun things).
If a balloon is inflated to 2 L at 26 ?C and then cooled in liquid nitrogen, what is the final volume
of the balloon?
ANSWER: Its Volume directly proportional to temperature
so V1 = k T1 and V2=k T2
so eliminating constants V1/T1=V2/T2
All temperatures must be in kelvin
2/299=V2 / Temperature of liquid nitrogen in Kelvin
V2=2/299 x 77
=0.52 litres
11.
ANSWER: 7.
Solubility of Ag+ S = (Ksp)^0.5=6.325*10^-7M
Normality N = Molarity M for AgBr as charge =1
Concentrations of Ag in a, b and c are much higher than solubility S.
So Ag+ concentration = S
page=-log(S)=6.2
8.
Reduction half cell
1/2O2 + 2e- = O^2Oxidation half cell
H2O+CO=CO2+2e- +2H+
So no of electrons n=2
G0=-263*10^3j/mol
E0=G0/(-n*96500)=1.36V
T=298K
R=8.314j/K mol
G0=-R*T*2.303*log(K)
Log(K)=-263*10^3/(-2.303*8.314*298)=46.1
K=1.24*10^46
15. General chemistry 2 question please help!
If the potential of hydrogen electrode based on the half-reaction
2 H+(aq) + 2e^- ----> H2(g)
is 0.000v at pH = 0.00, what is the potential of the same electrode at pH = 7.00?
ANSWER: from Nernest equation
E = E0 -RT/nF* log [product/[reactant]
Here R= universal gas constant 8.314 J/K mol
T = absolute temperature =25(0C)= 298k
F= faraday = 96500 Coloumb/mol
n = no of moles of electrons are transfered
RT/F= 0.0591
E= E0 -0.0591/n* log [product]/[reactant]
pH= 7
[H+]=10^-7 M
E= E0 -0.0591/n* log [1]/[H+]^2
E = 0- 0.0591/2 * log [1]/[10^-7]^2
E = -0.414 V
16. The chemistry of an ion exchange column is discussed in chapter 6-3 of the technique book.
If 10.00 mL of a solution containing 0.20 M NaCl and 0.25 M AlCl3 is introduced into a cation
exchange column, how many moles of hydronium ions will be released by the column?
ANSWER: First, you have to figure out how many moles of each salt you have.
molarity = moles/volume;
0.20 = moles NaCl/0.01 L
moles NaCl = 0.002
0.25 = moles AlCl3/0.01 L
moles AlCl3 = 0.0025
each mole of Na+ replaces 1 mole of H+
each mole of Al3+ replaces 3 moles of H+
0.002 mol + 3*0.0025 = 0.0095 mole H+
17. . How many grams of helium must be released to reduce the pressure to 69 atm assuming
ideal gas behavior. Note : 334-mL cylinder for use in chemistry lectures contains 5.209 g of
helium at 23 ∘C.
2. Calculate the density of NO2 gas at 0.940 atm and 36 ∘C.
3. Calculate the molar mass of a gas if 2.70 g occupies 0.865 L at 690 torr and 36 ∘C.
ANSWER: (1)We know that PV = nRT
PV = ( m/M) RT
Where
P = Pressure = 69 atm
V = Volume of cylinder = 334 mL = 0.334 L
m = mass of He = ?
M = Molar mass of He = 4 g/mol
R = gas constant = 0.0821 L-atm/(mol-K)
T = temperature = 23 oc = 23 + 273 = 296 K
Plug the values we get
m = (MPV) / (RT )
= ( 4x69x0.334) / ( 0.0821x296)
= 3.793 g
Therefore the mass of He at 69 atm is 3.793 g
But from the Lectures notes the mass of He is 5.209 g
So the mass of He released = 5.209 - 3.793 = 1.416 g
(2) We know that PV = (m/M) RT
PM = (m /V) RT
PM = dRT
Where
P = Pressure = 0.940 atm
M = Molar mass of NO2 = At.mass of N + (2xAt.mass of O )
= 14 + (2x16)
= 46 g/mol
d = density = ?
R = gas constant = 0.0821 Latm / (mol-K)
T = temperature = 36 oC = 36 + 273 = 309 K
Plug the values we get
d = (PM) / (RT)
= 1.709 g/L
(3) We have PV = (m/M) RT
Where
P = pressure = 690 torr
= 690 / 760 atm
Since 1 atm = 760 torr
= 0.908 atm
V = volume = 0.865 L
m = mas of gas = 2.70 g
M = molar mass of gas = ?
R = gas constant = 0.0821 L-atm / (mol-K)
T = Temperature = 36 oC = 36 + 273 = 309 K
Plug the values we get
M = (mRT) / (PV)
= ( 2.70 x 0.0821 x 309) / ( 0.908 x 0.865)
= 87.2 g/mol
18.
ANSWER: Given equation is C6H12O6 (aq) -------------> C2H5OH(l) + CO2 (g)
Balanced equation is C6H12O6 (aq) -------------> 2C2H5OH(l) + 2CO2 (g)
Given that density of ethanol (C2H5OH) = 0.789 g/mL
Molar mass of glucose ( C6H12O6)= 180 g/mol
Molar mass of ethanol (C2H5OH) = 46 g/mol
Theoritical yield of reaction:
Given that 325.0 of glucose yields 119.2 mL of ethanol.
C6H12O6 (aq) -------------> 2C2H5OH(l) + 2CO2 (g)
1 mol 2 mol
180 g 2x 46 g = 92 g
325g ?
Hence, 180 g of glucose produces 92 g of ethanol.
So, 325 g of glucose produces ? g of ethanol .
? = ( 325 g of glucose/180 g of glucose ) x 92 g of ethanol
= 166.1 g of ethanol
mass of ethanol = 166.1 g
This is called theoretical yield of the reaction.
Percent yiled of reaction:
Given that 325.0 of glucose yields 119.2 mL of ethanol.
mass of ethanol = volume x density = 119.2 mL x 0.789 g/mL = 94.05 g
Hence, actual yield of ethanol = 94.05 g
Then,
percent yield = (actual yield / theoretical yield) x 100
= (94.05 g of ethanol / 166.1) X 100
= 56.6 %
Percent yield = 56.6 %
19. the crc handbook of chemistry and physics gives the solubility of carbon dioxide (coming
from NaHCO3) in water at 25 degrees celsius as 0.145g in 100ml. assuming that the solubility of
carbon dioxide is the same in the acetic acid solution as in water, how much carbon dioxide (in
ml of gas at 25 degrees celsius and 1atm) might be dissolved in the 4ml used for each reaction
ANSWER: Solution :Solubility of the CO2 is 0.145 g / 100 ml
Therefore in 4 ml solution amount of the CO2 that can be soluble is calculated as
4 ml * 0.145 g / 100 ml = 0.0058 g
Now lets calculate the moles of the CO2
Moles of CO2 = 0.0058 g / 44.01 g per mol = 0.000132 mol
Now lets calculate the volume of the CO2 gas at the given conditions
T = 25 C +273 = 298 K
PV= nRT
V= nRT/P
= 0.000132 mol * 0.08206 L atm per mol K * 298 K / 1 atm
= 0.00322 L
0.00322 L * 1000 ml / 1 L = 3.22 ml
So in the 4 ml solution volume of the CO2 3.22 ml CO2
For per ml of the solution amount of the CO2 that can be soluble = 3.22 / 4 ml = 0.805 ml CO2
20. To determine whether a shiny gold-colored rock is actually gold, a chemistry student decides
to measure its heat capacity. She first weighs the rock and finds it has a mass of 4.5 g . She then
finds that upon absorption of 54.8 J of heat, the temperature of the rock rises from 25 ∘C to 60
∘C. Find the specific heat capacity of the substance composing the rock.
ANSWER: The amount of heat absorbed by rock , Q = mcdt
Where
m = mass of rock = 4.5 g
c = specific heat capacity of the substance composing the rock = ?
dt = change in temperature
= final - initial temperature
= 60 - 25
= 35 oC
Q = amount of Heat absorbed = 54.8 L
Plug the values we get
c = Q / ( mdt)
= 54.8/ (4.5 x35)
= 0.348 J/(goC)
Therefore the specific heat capacity of the substance composing the rock is 0.348 J/(goC)
21.
ANSWER: We are given, [NaOH] = 0.8500 M, , volume = 100.0 mL ,
[HOAc] = 0.8404 M, volume = 100.0 mL
Total volume = 100.0 +100.0 = 200.0 mL
We assume density of solution = 1.00 g/mL
So mass = 200.0 g , heat capacity = 4.184 J/goC.
Initial temp = 17.52 oC, final temp = 22.59oC
We know formula,
Heat, q = m *C*Δt
So, q = 200.0 g * 4.184 J/goC * ( 22.59 – 17.52)oC
= 4242.6 J
= 4.243 kJ
Heat loss = heat gain
Moles of NaOH = 0.100 L * 0.8500 M = 0.085 moles
Moles of HOAc = 0.100 L * 0.8404 M = 0.08404 moles
NaOH + HOAc -----> NaOAc + H2O
0.085
0.08404
0.08404
So limiting reactant is HOAc
So, heat = - ΔH
So, ΔHnut = - 4.243 kJ /0.08404 mole
= -50.5 kJ/mol
22. Occasionally students enjoy a cup of coffee or tea while studying chemistry. The amount of caffeine
in those beverages varies greatly. At a temperature of 65 degrees celsius, the maximum amount of
caffeine (C8H10N4O2) that can be dissolved is 455 g/L. What would be the maximum molarity of
caffeine in a 425 mL cup of 65 degree C coffee?
ANSWER: caffeine solubility = 455 g/L
for 1 Litre------------------> 455 g
425ml (0.425 L) ----------> 0.425 x 455 / 1 L = 193 .375 g /L
molar mass of caffeine = 194.19 g/mol
molarity = (solubility in g /L ) / molar mass
= (193.375 / 194.19)
= 0.996 M
23. Chapter is over Equilibrium Chemistry and connecting to Gravimetric systematic solving.
6. Write a charge balance equation and mass balance equations for the following solutions. Some
solutions may have more than one mass balance
Please help I don't understand how to derive these.
a. 0.1 M NaCL
b. 0.1 M HCl
c. 0.1 M HF
g. 0.10 M HCl and 0.05 M NaNO2
ANSWER: a.
NaCl (strong electrolyte) is found in solution as Na+(aq) 0.1 M and Cl-(aq) 0.1 M. No one react
with water.
Mass balance :
CNaCl (formal) = ([Na+] + [Cl-])/2 or
NaCl mass = mass of Na+ + mass of Cl-
Charge balance:
(+1)[Na+] + (-1)[Cl-] = 0
b.
HCl is a strong acid. It react completely with water.
HCl + H2O = H3O+ + ClMass balance :
CHCl (formal) = ([H3O+] + [Cl-])/2
Charge balance:
(+1)[H3O+] + (-1)[Cl-] = 0
You can add also the equilibrium
2H2O = H3O+ + ClAnd write
(+1)[H3O+]total + (-1)[Cl-] + (-1)[HO-] = 0
Or [H3O+]total = [Cl-] + [HO-]
(but [HO-] is only 1x10-13M, a negligible value)
c.
HF is a weak acid (Ka = 7.2x10-4 , pKa = 3.14)
[H+] = [F-] = (Ka.Cacid)1/2 = 8.5 x10-3M
Mass balance :
CHF (formal) = [HF] + ([F-] + [H+])/2
Charge balance:
[H+] = [F-]
Or considering also the dissociation of H2O
[H+]total = [F-] +[HO-]
d.
NaNO2 (strong electrolyte) is found in solution as Na+(aq) 0.1 M and NO2-(aq) 0.1 M. NO2- react
with H+ (from HCl) and is completely neutralized. So the solution will contains:
Na+(aq) 0.1 M and Cl-(aq) 0.1 M and the weak acid HNO2 that dissociates (see b.)
Mass balance :
CNaNO2 + CHCl = ([Na+] + [Cl-] ) /2 + [HNO2] + [NO2-]
Charge balance
[H+] + [Na+] = [Cl-] + [NO2-]
Or considering also the dissociation of H2O
[H+]total + [Na+] = [Cl-] + [NO2-] + [HO-]
24. Chapter is over Equilibrium Chemistry and connecting to Gravimetric systematic solving.
6. Write a charge balance equation and mass balance equations for the following solutions. Some
solutions may have more than one mass balance
Please help I don't understand how to derive these.
a. 0.1 M NaCL
b. 0.1 M HCl
c. 0.1 M HF
g. 0.10 M HCl and 0.05 M NaNO2
ANSWER: a) . 0.1 M NaCl
mass balance
NaCl ------------------> Na+ + Cl[NaCl] = [Na+ ] + [Cl-] = 0.1 + 0.1 = 0.2
charge balance :
mole of Na+ = moles of Clso charge balacnce = 0.1
(b)
mass balance
HCl ------------------> H+ + Cl[HCl] = [H+ ] + [Cl-] = 0.1 + 0.1 = 0.2
charge balance :
mole of H+ = moles of Clso charge balacnce = 0.1
c)
HF + H2O ---------------> H3O + Ftotal ions
H+ , F- , H+ , OHmass balance = [H+] = [F-] + [OH-] = 0.1 + 0.1 = 0.2
charge balance = 0.1
charge balance [Na+] = [Cl-]
25.
ANSWER: m = 1.58 g of PbCl2
V = 36.39 ml
M = 1.65 M Hcl
V = 234.5 ml
M = 0.3445 Pb(NO3)2
% yield?
given
2HCl + Pb(NO3)2 --> PBCl + 2HNO3
%yield = real value / theoretical value * 100%
real value = 1.58 g of PbCl2
theoretical value:
mol of HCl = M*V = 1.65*36.39 = 60.04 mmol of Acid
mol of PB(NO3)2 = M*V = 0.3445*234.5 = 80.8 mmol of Pb(NO3)2
there is clearly HCl limiting the reaction (60 mmol of Acid will react only with 30 mmol of
PB...)
Therefore:
2HCl: PbCl2
Ratio is 1/2*60.04 mmol = 30.02 mmol of PBCl2 is expeted to precipitate
MW of PbCl2 = 278.1 g/mol
mass = mol*MW = (30.02*10^-3)(278.1) = 8.345 grams of PbCl2 could be produced
%yield = 1.58/8.345 *100 = 18.933%
26. A friend purchases a helium‐filled balloon and notices that the balloon is made of a foil‐type
material (called Mylar). Your friend knows you are taking chemistry and asks you why the
helium balloon is made of a different material than regular balloons blown up with air. (For this
example, simplify by assuming air used in regular balloons is 100% N2.) Give your friend an
answer in words supported by 1 appropriate calculation.
ANSWER: Solution :Lets assume we have 1 mole of the each gas in the ballon at the STP conditions
at STP condition 1 mole gas = 22.4 L
so both balloons will have volume 22.4 L
but the denities of the gases are different
because mass of 1 mol N2 = 28.014 g per mol
and molar mass of He = 4.0026 g per mol
so the density of the N2 gas = 28.014 g / 22.4 L = 1.25 g per / L
and density of the Helium = 4.0026 g / 22.4 L = 0.179 g / L
Therefore as the when the balloon is filled with the helium gas then it have very lower density
which can float in the air
And when the temperature of the atmosphere increase the pressure inside the helium balloon
increases which causes the increase in the volume of the helium balloon therefore it can blow up.
therefore the material used to make the Helium balloon is foil type material which used to
prevent the
27. Chapter is over Equilibrium Chemistry and connecting to Gravimetric systematic solving.
6. Write a charge balance equation and mass balance equations for the following solutions. Some
solutions may have more than one mass balance
Please help I don't understand how to derive these.
a. 0.1 M NaCL
b. 0.1 M HCl
c. 0.1 M HF
g. 0.10 M HCl and 0.05 M NaNO2
ANSWER: It is to be noted that in a charge balance equation
0.1 M NaCl:
H2O ↔ H+ + OHNaCl ↔ Na+ + ClMass balance equation:
0.1 [Na+] = 0.1 [Cl-]
Charge balance equation:
[H+]+[Na+]=[OH-]+[Cl-]
0.1 M HCl:
H2O ↔ H+ + OHHCl ↔ H+ + ClMass balance equation:
0.1 [H+] = 0.1 [Cl-]
Charge balance equation:
[H+] =[OH-]+[Cl-]
0.1 M HF:
H2O ↔ H+ + OHHF ↔ H+ + FMass balance equation:
0.1 [H+] = 0.1 [F-]
Charge balance equation:
[H+]=[OH-]+[F-]
0.1 M HCl and 0.05 NaNO2:
H2O ↔ H+ + OHHCl ↔ H+ + ClNaNO2 ↔ Na+ + NO2Mass balance equation:
0..05 [Na+] = 0..05 [NO2-]
Charge balance equation:
[H+]+[Na+]=[OH-]+[NO2-] +[Cl-]
28. A friend purchases a helium‐filled balloon and notices that the balloon is made of a foil‐type material
(called Mylar). Your friend knows you are taking chemistry and asks you why the helium balloon is made
of a different material than regular balloons blown up with air. (For this example, simplify by assuming
air used in regular balloons is 100% N2.) Give your friend an answer in words supported by 1
appropriate calculation
ANSWER: Since the molecular mass of He is smaller than that of N2, He has high rate of
diffusion/effusion in comparison to N2. This makes it easier for He to escape from the baloon.
This can be explained from the following calculation.
According to Grahm's law the rate of effusion/diffusion of a gas is inversely proportional to
square root of molecular mass.
Molecular mass of N2 = 28 g/mol
Mlecular mass of He = 4g/mol
Suppose the rate of diffusion/effusion of N2 and He are respectively r1 and r2, then according to
Grahm's law of diffusion
r1/r2 = underroot (4 / 28) = 0.378
=> r2 = 2.65xr1
Hence the rate of diffusion of He is 2.65 times higher than that of N2.
Hence to prevent its escape He is filled in a foil - type material like mylar that has very small
pore size and makes it difficult for He to escape.
29. The reaction A2(g) <----> 2 A(g) takes place in a closed reaction vessel at constant temperature and
volume. Initially, the reaction vessel only contains A2(g) at a pressure of 1.28 atm. After the reaction
reaches equilibrium, the total pressure in the reaction vessel is 2.5 atm. What is the value of the
equilibrium constant, Kp, for the reaction?
ANSWER: the reaction is
A2 ---> 2A
now
using ICE table
initial presure of A2 , A are 1.28 , 0
change in pressure of A2 , A are -x , + 2x
equilibrium pressure of A2 , A are 1.28 - x , 2x
given
total pressure at equilibrium = 2.5
so
1.28- x + 2x = 2.5
x = 1.22
so
pA = 2x = 2 * 1.22 = 2.44
pA2 = 1.28 - x = 1.28 - 1.22 = 0.06
now
consider the reaction
A2 ----> 2 A
the equilibrium constant is given by
Kp = [A]^2 / [A2]
Kp = [2.44]^2 / [0.06]
Kp = 99.23
so
the equilibrium constant is 99.23
30. i havent take it chemistry for a long time
so i forgot how to do the caculation of the formulate
can someone help me and show me how to do it thank you
calculate how many g of ammonia will be produce from 9.35 g ofnitrogen gas and excess
hydrogen using the following equation
N2(g) + 3 H2(g) --------------> 2NH3(g)
ANSWER: First calculate the moles of nitrogen gas you have:
moles N2 = mass / molar mass ofN2
= 9.35 g / (14.01*2) g.mol-1
= 0.3337 mol
The equation is balanced already, and it tells us that 1 mole ofnitrogen gas (N2) react to form 2 moles of
ammonia(NH3). Therefore we can calculate the moles of ammoniaproduced:
moles NH3 = 2*moles N2
= 2*0.3337
= 0.6674 mol
Now do the reverse of the first calculation to find the mass ofammonia produced:
mass = moles * molar mass of NH3
= 0.6674 mol * (14.01 + 1.008*3) g.mol-1
= 11.4 g NH3
31. It's
a general chemistry here's the awesome question! a30.0 g ice cube at -10.0 degrees
Celsius/centigrade is added to100ml water at 22.2 degrees delsius in insulated container. Afterice
melts, what will be final tempterature?
ANSWER:. Things you need to know for this question:
q =m*c*ΔT
(q = heat gain, -q= heat loss, m = mass of system, c = heat capacity, T =temperature)
c(H2O liquid) = 4184J.kg-1.K-1
enthalphy(latent heat) of fusion for water (Hf) = 333.35kJ/kg
Therefore the heat energy given out by the ice is:
q = -(heat given out while melting + heat givenout while at melting point)
= -(m(ice)*c(water)*(Tf - Ti)+ heat of fusion * m(ice))
= -(0.030 kg * 4.184 kJ/kg.K *(Tf - 273K))
= -(0.12552 kJ/K * (Tf - 273K))
= -0.12552*Tf + 34.2670
Meanwhile, the heat taken in by the water is:
q = m(water)*c(water)*(Tf - Ti)
= 0.1 kg * 4.184 kJ/kg.K * (Tf- 295.2K)
= 0.4184 (Tf - 295.2)
= 0.4184*Tf - 123.5117
Equating the two sides:
-0.1255*Tf + 34.2670 = 0.4184*Tf - 123.5117
0.5439*Tf = 157.7787
Tf = 290.1 K = 17.1 oC
32. Perspiration has a slight acidity because of the presence oflactic acid (C3H6O3). Suppose weanalyze
the perspiration of a chemistry student running late toclass. If the density of the perspiration were 1.23
g/mL and the mass percent lactic acid were foundto be 4.82%, what would you calculate asthe mole
fraction, molarity, and molality of the lactic acid?(Assume that the perspiration contains only lactic acid
andwater.)
ANSWER: We Know that :
Xlacticacid = number of moles of lactic acid/ Total number of moles of solution
= 4.82 gm / 90 g / mole / 4.82 gm/ 90 g / mole + 95.18 gm / 108 gm/mole
= 0.05728
Molarity of Lacticacid = Number of moles of Lactic acid / Volume of the solution L
= [ 4.82 g / 90 g / mole / weight of thesubstance / density ( ml ) ] x
1000
= [ 4.82 g / 90 g / mole / 95.18 g/ 1.23 g / ml ] * 1000
= 0.6920 M
Molality ofthe Lactic acid = number of moles of lacticacid / weight of the solvent Kg
= 0.05355 / 95.18 g / 1000
= 0.56267 molal.
Perspirationcontains ortho- cresol or para-cresol as thesolvent.
33. Electric power is typically stated in units of watts (1watt = 1 J/s). About 95% of the power output of
anincandescent light bulb is converted to heat and 5% tolight. If 10% of that light shines on your
chemistry text, howmany photons per second shine on the book from a 75-wattbulb? Assume the
photons have a wavelength of 550 nm.
ANSWER: From a 75watt bulb 5% is converted to light
75watt = 75J/s
then 5% of 75J/s = 3.75J/s
then 10% of this light shines = 3.75J/s * 10/100= 0.375J/s
Calculate the amount of energy when λ is 550nm for1photon
E = hc/λ = 6.626*10-34J.s *3*108 m/s / 550*10-9 m
=3.6*10-19 J
To calculate the numbe of photon stricking per sec is= 3.75J/s / 3.6*10-19 J
= 1.037*1018 photons /s
34. Calculate Ka for each of the following acids, given its pKa. Rank the compounds in order of
decreasing acidity. Please explain in simple terms how to do this work.
(a) Aspirin: pKa = 3.48
(b) Vitamin C (ascorbic acid): pKa = 4.17
(c) Formic acid (present in sting of ants): pKa = 3.75
(d) Oxalic acid (poisonous substance found in certain berries): pKa = 1.19
The book has no examples this is organic chemistry 1 edition 8
ANSWER: We know pKa = - log Ka
--- Ka = 10^-pKa
(a) Aspirin: pKa = 3.48
Ka = 10^-pKa
= 10^-3.48
= 3.31*10^-4
(b) Vitamin C (ascorbic acid): pKa = 4.17
Ka = 10^-pKa
= 10^-4.17
= 6.76*10^-5
(c) Formic acid (present in sting of ants): pKa = 3.75
Ka = 10^-pKa
= 10^-3.75
= 1.78*10^-4
(d) Oxalic acid (poisonous substance found in certain berries): pKa = 1.19
Ka = 10^-pKa
= 10^-1.19
= 64.56*10^-3
---------------------------------------------------------------------As the Ka value is more more is its acidic strength
( Ka is nothing but acid dissociation constant .As the dissociation isd more more is its acidic
nature )
So the order of acidity is :
Oxalic acid > Aspirin > formic acid > Vitamin C
35. A professor wanted to set up a similar experiment as the one you performed in lab. The
professor wanted to use Al(OH)3 in place of Ca(OH)2. Calculate how many mL of saturated
Al(OH)3 solution it would take to titrate against 12.00mL of 0.0542M HCl solution? The Ksp of
Al(OH)3 is 3.0x10-34. Do you think this would be reasonable experiment for a general
chemistry lab?
ANSWER: 12.00mL * 0.0542M HCl = 0.6504mmols H+ = 0.6504 mmols OHKsp = [Al][OH]^3
3.0E-34 = x*(3x)^3 = 27x^4
x^4=1.1* 10^-34
X= 3.239 * 10^-9
[OH-]=3x= 9.717 * 10^-9
(12.00mL * 0.0542) = ( [OH-] * V Al(OH)3 )
(12.00mL*0.0542)/(9.717 * 10^-9)=V
6.67 * 10^7 ml of Al(OH3)
36. Find the mass of sodium formate that must be dissolved in 150.0cm3 of a 1.2M solution of
formic acid to prepare a buffer solution with pH = 3.60.
Please help with this. I know there are similar questions posted already, but I cannot get the math
to work out for some reason and I only have one more chance on MasteringChemistry. TIA
ANSWER: mass of NaFormate
V1 = 150 cm3 that is 150 ml
M1 = 1.2 Formic Acid
pH of buffer = 3.6
Sine it is a Buffer, let us model this iwth the Henderson-Hasselbach Equation
pH = pKa + log([common ion][acid])
pKa = 3.75 ***From internet databases
[Acid] = 1.2M
3.60 = 3.75 + log([salt]/1.2)
Solve for [Salt]
-0.15 = log ([salt]/1.2)
10^-.15 = [salt] /1.2
0.70794 * 1.2 = [salt]
[salt] = 0.8495
But we need mass not concentration!
[salt] = moles of salt / L solution
0.8495 = moels of salt / (0.150 L)
moles of salt = 0.8495*.15 = 0.1274 moles of sodium formate
but we need mss, not moles
MW of NaFormate = 68.01 g/gmol
mass = moles*MW = 0.1274 mol * 68.01 g/gmol = 8.6644 grams of Sodium Formate
mass = 8.6644 grams of Sodium Formate
37.
ANSWER: Given, 50.0 mL of 0.100 M HClO
HClO <-------------> H+ + ClOI 0.1 0 0
C -x +x +x
E 0.1 - x +x +x
Ka = [H+][ClO-] / [HClO]
3.5 x 10^-8 = x^2 / 0.1-x
x^2 = 3.5 x 10^-9
x = 5.92 x 10^-5 M = [H+]
pH = -log[H+] = -log[5.92 x 10^-5] = 4.23
(a)
Thus, pH before addition of any KOH, pH = 4.23
---------------------------------------------------------------------------------------------------------------(b)
Moles of HClO = Molarity x Volume(in L) = 0.100 x 0.050 = 5 x 10^-3 mol
Moles of KOH = Molarity x Volume(in L) = 0.100 x 0.025 = 2.5 x 10^-3 mol
Moles of HClO left over = (5 x 10^-3 mol) - (2.5 x 10^-3 mol) = 2.5 x 10^-3 mol
Total volume = 50.0 + 25.0 = 75.0 mL = 0.075 L
[HClO] = 2.5 x 10^-3 mol / 0.075 = 0.033 M
HClO <-------------> H+ + ClOI 0.033 0 0
C -x +x +x
E 0.033-x +x +x
Ka = [H+][ClO-] / [HClO]
3.5 x 10^-8 = x^2 / 0.033-x
x^2 = 1.167 x 10^-9
x = 3.42 x 10^-5 M = [H+]
pH = -log[H+] = -log[3.42 x 10^-5] = 4.46
pH = 4.46
--------------------------------------------------------------------------------------------------------------------(c)
Moles of HClO = Molarity x Volume(in L) = 0.100 x 0.050 = 5 x 10^-3 mol
Moles of KOH = Molarity x Volume(in L) = 0.100 x 0.040 = 4.0 x 10^-3 mol
Moles of HClO left over = (5 x 10^-3 mol) - (4.0 x 10^-3 mol) = 1.0 x 10^-3 mol
Total volume = 50.0 + 40.0 = 90.0 mL = 0.090 L
[HClO] = 1.0 x 10^-3 mol / 0.090 = 0.011 M
HClO <-------------> H+ + ClOI 0.011 0 0
C -x +x +x
E 0.011-x +x +x
Ka = [H+][ClO-] / [HClO]
3.5 x 10^-8 = x^2 / 0.011-x
x^2 = 3.89 x 10^-10
x = 1.97 x 10^-5 M = [H+]
pH = -log[H+] = -log[1.97 x 10^-5] = 4.70
pH = 4.70
----------------------------------------------------------------------------------------------------------------------(d) after addition of 50.0 mL of KOH
Moles of HClO = Molarity x Volume(in L) = 0.100 x 0.050 = 5 x 10^-3 mol
Moles of KOH = Molarity x Volume(in L) = 0.100 x 0.050 = 5.0 x 10^-3 mol
Moles of KClOformed = 5.0 x 10^-3 mol
Total volume = 50.0 + 50.0 = 100.0 mL = 0.100 L
[KClO] = 5.0 x 10^-3 mol / 0.100 = 0.05 M
KClO + H2O <-------------> HClO + OHI 0.05 0 0
C -x +x +x
E 0.05-x +x +x
Kb = [OH-][HClO] / [KClO]
Kb = Kw / Ka = (1 x 10^-14) / (3.5 x 10^-8) = 2.86 x 10^-7
2.86 x 10^-7 = x^2 / 0.05-x
x^2 = 1.43 x 10^-8
x = 1.19 x 10^-4 M = [OH-]
pOH = -log[OH-] = -log[1.19 x 10^-4] = 3.92
pH = 14 - 3.92 = 10.07
--------------------------------------------------------------------------------------------------------------------------(e) after addition of 60.0 mL of KOH
Moles of HClO = Molarity x Volume(in L) = 0.100 x 0.050 = 5 x 10^-3 mol
Moles of KOH = Molarity x Volume(in L) = 0.100 x 0.060 = 6.0 x 10^-3 mol
Moles of KOH left over = (6.0 x 10^-3) - (5.0 x 10^-3) = 1.0 x 10^-3 mol
Total volume = 50.0 + 60.0 = 110.0 mL = 0.110 L
[KOH] = 1.0 x 10^-3 mol / 0.110 = 0.0091 M
As, KOH is a strong base,
pOH = -log [OH-]
pOH = -log[0.0091]
pOH = 2.04
pH = 14 - 2.04 = 11.95
pH = 11.95
38. In chemistry, 6.022Ý 1023 is an important number called the Avogadro constant (or
Avogadro\'s number). It expresses the number of atoms or molecules in a mole. Notice that it is a
really, really big number, as indicated by scientific notation with a large positive exponent. Enter
6.022Ý 1023 into the calculator shown here. Then, by trial and error, find the largest number that
can be added to 6.022Ý 1023 without changing its displayed value
ANSWER: It would be 4.99999999 x 1013
As you add 6.022 x 1023 + 4.99999999 x 1013 = 6.022 x 1023
As you add 6.022 x 1023 + 5 x 1013 = 6.022000001 x 1023
My display changed to 6.022000001 x 1023 as I changed the value to 5 x 1013
So the largest number you can add to Avagadro number without changing its value is
4.99999999 x 1013
39.
ANSWER: a) The atomic number of oxygen-18 is 8. It's mass number is 18.
One atom of oxygen-18 contains 8 protons (the number of protons is equal to the atomic number
8).
The number of neutrons present in one atom of oxygen -18 is equal the difference between mass
number and atomic number.
It is
.
Thus one atom of oxygen-18 contains 10 neutrons.
b) When an atom is electrically neutral (has no overall (net) charge) then the number of protons
is equal to the number of electrons it contains.
Each proton has unit positive charge and each electron has unit negative charge. When the
number of protons is equal to the number of electrons, the total number of positive charges is
equal to the total number of negative charges. Hence, the atom is electrically neutral.
c) The number of electrons present in one atom of oxygen-18 is equal to the number of protons.
It is equal to 8.
40. During a chemistry experiment, the cork in
a 0.4-foot tall beaker with an effervescent solution pops off with an initial velocity of 15feet per second.
The quadratic model for the
situation is h = -16t2 + vt + s
where v is the initial velocity, t is in seconds, and s is the initial height.
How many seconds does it take for the cork to hit the table?
ANSWER: Substitute
0 = -16t^2 + 15t + .4
The quadratic formula for an equation ax^2 + bx + c = 0 is
x=-b±√(b^2 - 4ac) / 2a
so substitute again
t= -15 ± √(225 - 4(-16)(.4)) / 2(-16)
Since the + will turn out to be -.0259 seconds, we have to do
t= -15 - √(225 - 4(-16)(.4)) / 2(-16)
= .963 seconds is what it should be for the equation
41. What is the boiling point of a solution containing 0.250g of copper (ii) chloride, an ionic
compound, in 100.0g of water? The molal freezing point constant for water is 1.858 C/m, and the
molal boiling point constant for 0.512 C/m.
ANSWER: DTb= i* K* m
i= 3 for CuCL2
Kb= 0.512 C/m
molality (m) = 0.25/ 134.45*1000/100 = 0.0186 molal
TS-Tsol = 3*0.512*0.0186
Tsol = boiling point of water
Ts -100 = 0.0285696
Ts = Boiling point of Solution = 100 + 0.0285696 = 100.0285696 C
42. Compare an average oxidation state (look this up in your organic chemistry book) of all
carbons of the first reactant and the last product of the glycolysis pathway. Compare an average
oxidation state of all carbons in the products of homolactic fermentation with that of glucose.
Where does glycolytic oxidative potential come from? In order to continue, this potential has to
be regenerated. How is this achieved? Relate this to the bread and wine making.
ANSWER: 1. In glycolysis, the first reactant is glucose (C6H12O6) and the last product is
pyruvic acid (CH3(C=O)COOH).
oxidation state of carbon in glucose: H carries +1 and O carries -2
therefore, oxidation state of carbon = 12(+1) + 6 (-2)
= +12-12 = 0
oxidation state of carbon in pyruvic acid, CH3(C=O)COOH or C3H4O3
there are three carbons, 1st carbon, CH3 = Oxidation state = -3
2nd carbon, C=O, Oxidation state = +2
3rd carbon, COOH, Oxidation state = +3
average oxidation state of carbon= 0+ (-3) + (+3) + (+2) = +2
2. oxidation state of carbon in homolactic fermentation to glucose, here glucose is converted to
lactic acid.
glucose C6H12O6 = 0
C3H6O3 = 6 (+1) + 3 (-2) = 0
3. Therefore the average oxidation state in glycolysis is +2 of carbon and that in homolactic
fermentation of glucose is 0.
4. Glycolytic oxidative potential is achieved by the tranfer of hydrogen from glucose to NAD
forming NADH, this way the reaction is facilitated. (oxitation potential is the measure of an
element to oxidize or lose electrons). in Wine making alcoholic fermentation takes place by
glycolysis and in bread making also glycolysis takes place in a similar way.
44. Chemistry questions please help!
1)calculate the number of grams of silver chromate formed from 25.0 ml of 0.125 M silver
nitrate and 15.0 ml of 0.120 M Potassium chromate .
2)how many ml of 12.0 M hydrochloric acid are needed to make 2.50 L of 0.283 M acid
3)What is the molarity of a sulfuric acid solution if 25.0 lm required 38.2 ml of 0.135 M sodium
hydroxide to neutralize it?
ANSWER: moles of AgNO3 = 25.*0.125 = 3.125 milimoles
mole sof K2 CrO4 = 15*0.120 = 1.80 milimoles.
reaction is .
2AgNO3+K2CrO4=>2KNO3+Ag2CrO4
so ,
limiting ragent is AgNO3 ,
so ,
moles of Ag2 CrO4 formed = 3.125/2 = 1.5625 milimoles
so mass ,= 1.5625 * 331.73
mass = 518.328 mg = 0.518 g
2. so ,
M1*V1 = M2 *V2 ,
12 * V = 2.5*0.283 ,
V = 0.5896 L ~ 58.96 mL
3.
for H2 SO4 ( M1,V1) and NaOH ( M2,V2)
2 ( M1V1 ) = M2*V2 ,
2*25 * M 1 = 38.2*0.135 ,
M1 = 0.10314 moles per Litre
so ,
Molarity of H2SO4 = 0.10314 M
45. Calculate (i) the change in the melting point and (ii) the change in the boiling point when a
solution of a 1:1 electrolyte of molar mass 88.7 g mol-1 is prepared by adding 65.2 g of the
electrolyte to 514.9 g of a polar solvent. For the solvent Kf = 0.868 K kg mol-1and Kb = 2.290 K
kg mol-1. Include signs with your answers.
ANSWER: 1)
DTf = Kf *molality
molality = wt/molwt(1000/wt of solvent)
molality = 65.2/88.7(1000/514.9)
molality = 1.43 molal
DTb = i*Kb*m
= 2*2.290*1.43
DTb = 6.54
2)
DTf = i*Kf*m
DTf = 2*0.868*1.43
DTf = 2.482
46.
ANSWER: Question 1 )
M1 = 1.0 M
V1 = 2.5 ml
M2 = ?
V2 = 2.5 + 1.5 = 4.0 ml
M1 V1 = M2 V2
1.0 x 2.5 = M2 x 4.0
M2 = 0.625 M
Question 2 )
M1 = 1.0 M
V1 = 1.5 ml
M2 = ?
V2 = 1.5 + 2.5 = 4.0 ml
M1 V1 = M2 V2
1.0 x 1.5 = M2 x 4.0
M2 = 0.375 M
47.
ANSWER: For an isothermal process,
dU = dH = 0
work done is calculated as using equation,
dw = Vi-nRTln(Vf-nb)/(Vi-nb)-n^2.a(1/Vf-1/Vi)
= 35.50-3.25 x 8.314 x 375 ln(75.25-3.25 x 0.051)/(35.50-3.25 x 0.051)3.25^2.4.250(1/75.25-1/35.50)
= -7600.85 = -7.60 kJ
dq = -w = 7.60 kJ
dS = q/T = 7.60/375 = 0.022 kJ/K
dG = dH - TdS = -375 x 0.022 = -8.25 kJ
48.
ANSWER: 1. d. 2NaCl + H2SO4 = 2HCl + Na2SO4
e. Molar mass of NaCl = 58.44 g
Molar mass of HCl = 36.46 g
Molar mass of Na2SO4 = 124.04 g
Molar mass of H2SO4 = 98.079 g
2 mol or 58.44*2= 116.88 g NaCl reacts completely with 98.079 g H2SO4
So, 35 g NaCl reacts completely with = 98.079 * 35/116.88 g H2SO4
= 29.37 g H2SO4
f. 116.88 g NaCl reacts to completely to give 2 * 36.46 = 72.92 g HCl
So, 35 g NaCl reacts to completely to give = 72.92*35/116.88 g HCl
= 21.84 g HCl
% of yield = (8.6/21.84) * 100 = 39.4 %
g. When 72.92 g HCl forms, 124.04 g Na2SO4 forms
When 8.6 g HCl forms, 124.04*8.6/72.92 or 14.63 g Na2SO4 forms
Ans: 14.63 g Na2SO4
2. a. Balanced equation: B2H6 + 6 H2O = 2 H3BO3 + 6 H2
Molar mass of B2H6 = 27.67 g
Molar mass of H3BO3 = 61.83 g
27.67 g B2H6 reacts to give 123.66 g H3BO3
So, 28.16 g B2H6 reacts to give = 123.66 * 28.16/27.67= 125.85 g H3BO3
3.a.
2Ca + O2 = 2CaO
b. Molar mass of Ca = 40.078 g
Molar mass of CaO= 56.07 g
80.156 g Ca gives 112.14 g CaO
So, 5.64 g Ca gives = 112.14 * 5.64/80.156 = 7.89 g CaO = 7.89/56.07 mol of CaO = 0.14 mol of CaO
49. A chemistry student in lab needs to fill a temperature-control tank with water. The tank
measures 34.0cm long by 24.0cm wide by 12.0cm deep.In addition, as shown in the sketch
below, the student needs to allow 2.0cm between the top of the tank and the top of the water, and
a round-bottom flask with a diameter of 9.5cmwill be just barely submerged in the water.
Calculate the volume of water in liters which the student needs. Round your answer to the
nearest
0.1L
ANSWER: without any diagram or sketch I can only do this mathematically:
Volume: Long x Wide x Deep (all in dm, this will make that the answer results in Liters)
Volume: 3.4 x 2.4 x 1.0
Volume: 8.16 Liters. the flask does not involve in the problem if you do not have its height
50. A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully
weights out 305.mg of oxalic acid(H2C2O4), a diprotic acid that can be purchased inexpensively
in high purity, and dissolves it in 250.mL of distilled water. The student then titrates the oxalic
acid solution with her sodium hydroxide solution. When the titration reaches the equivalence
point, the student finds she has used 69.4mL of sodium hydroxide solution. Calculate the
molarity of the student's sodium hydroxide solution. Be sure your answer has the correct number
of significant digits.
ANSWER: Given mass of oxalic acid is 305 mg = 305x10-3 g
Molar mass of oxalic acid , H2C2O4 = (2x1) + (2x12)+(4x16) = 90 g/mol
So number of moles of oxalic acid , n = mass/ molar mass
= (305x10-3 g )/ 90 (g/mol)
= 3.39x10-3 mol
Given volume of oxalic acid solution , V = 250.0 mL = 0.250 L
Therefore Molarity of thre oxalic acid solution , M = Number of moles / Volume of solution in L
= (3.39x10-3 mol) / 0.250 L
= 0.0135 M
The balanced reaction between oxalic acid & NaOH is
H2C2O4 + 2NaOH
Na2C2O4 + 2H2O
According to the above balanced equation ,
1 mole of oxalic acid reacts with 2 moles of NaOH
3.39x10-3 mol of oxalic acid reacts with 2x3.39x10-3 mol= 6.78x10-3 moles of NaOH
So Molarity of the unknown NaOH solution , M = number of moles of NaOH reacted / Volume
of NaOH solution in L
= 6.78x10-3 mol / (69.4 /1000) L
= 0.098 M
Therefore the molarity of unknown NaOH solution = 0.098 M
51.
ANSWER: CaSO4 is more soluble ,so [SO42-] is due to dissociation of CaSO4.
Ksp=[Ca2+][SO42-]=2.4*10^-5
[Ca2+]=0.100M
2.4*10^-5=(0.100M) [SO42-]
[SO42-]=2.4*10^-5/0.100=24*10^-5
At [SO42-]=24*10^-5 ,[Ba2+] can be calculated
Ksp=[Ba2+][SO42-]
1.1*10^-10=[Ba2+]*(24*10^-5)
[Ba2+]=1.1*10^-10/(24*10^-5)=0.046*10^-5M
[Ba2+]=0.046*10^-5M
Fraction of Ba2+ precipitated=[Ba2+]/[Ba2+] initial =0.080M/(0.046*10^-5M)=1.74*10^-5
Fraction of Ba2+ precipitated=[Ba2+]/[Ba2+] initial =1.74*10^-5
52.
ANSWER: No of mole of NaHCO3 = 1 mol
No of mole of Na2CO3 = 1.5 mol
pka of HCO3- = 10.32
pH = pka +log(salt/acid)
= 10.32+log(1.5/1)
pH = 10.49
10. No of mol of HNO3 = 0.2*1 = 0.2 mol
pH = pka +log(salt-acid/acid+acid)
= 10.32+log((1.5-0.2)/(1+0.2))
= 10.354
53.
ASNWER: In order to solve such a problem we have to subtract/add the given reactins in such a
way so that the net result will be the reacion of our concern as explained. If we subtract reation 1
from reaction 3 we will get reaction 3(The reaction of our concern)
subtract (1) reaction from (2) reaction
Cl(g) + 2O2 (g) ------------> ClO(g) + O3(g) ΔH = 122.8KJ (HINT: When we subtract a
reaction from other we simply revrse it and change the sign of ΔH)
2O3(g) ---------------> 3O2(g) ΔH = -285.3KJ
So the net reaction
Cl(g) + O3 (g) ------------> ClO(g) + O2(g)
ΔH = ΔH1 - ΔH2
ΔH = -285.3 - (-122.8) = -162.5KJ
54. What is the distance between lines on a diffraction grating that produces a second-order maximum
for 790 nm red light at an angle of 61.0°?
ANSWER:. = wavelength = 790 x 10-9 m
= angle = 61
n=2
d = width of slit
Using the formula
d Sin = (2n + 1)
/2
d Sin61 = (5/2) (790 x 10-9 )
d = 2.26 x 10-6 m
55. Juan makes a measurement in a chemistry laboratory and records the result in his lab report.
Suppose that if Juan makes this measurement repeatedly, the standard deviation of his
measurements will be σ = 6 milligrams. Juan repeats the measurement four times and records the
mean x of his four measurements.
(a) What is the standard deviation of Juan's mean result? (That is, if Juan kept on making four
measurements and averaging them, what would be the standard deviation of all his x values?)
_________mg
(b) How many times must Juan repeat the measurement to reduce the standard deviation of x to
2?
ANSWER: Juan makes a measurement in a chemistry laboratory and records the result in his lab
report. Suppose that if Juan makes this measurement repeatedly, the standard deviation of his
measurements will be
σ = 6 milligrams
repeatation of measurements (n) = 4
a) What is the standard deviation of Juan's mean result?
Juan's mean result standard deviation will be,
σ / sqrt(n) = 6 / sqrt(4) = 3 milligrams
b) Given that σ = 6 xbar= 2 n = ?
How many times must Juan repeat the measurement to reduce the standard deviation of x to 2?
2 / 1 = 6 / sqrt(n)
sqrt(n) = 6 / 2
sqrt(n) = 6/2= 3
sqrt(n) = 3
n = (3)2 = 9
n=9
56.
ANSWER: We are given combustion reaction
Reaction –
C6H6(l) + O2(g) ----> H2O(g) + CO2(g)
First we need count each element from both side –
’
C6H6(l) + O2(g) ----> H2O(g) + CO2(g)
C=6
C=1
H=6
H=2
O=2
O=3
Now we need to balance the c first- In the reactant there are 6 C and in the product there are 1 C,
so we need to 6 C in product side also, so we need to put 6 CO2
C6H6(l) + O2(g) ----> H2O(g) + 6 CO2(g)
C=6
C=6
H=6
H=2
O=2
O = 13
Now we need to balance the H- In the reactant there are 6 H and in the product there are 2 H, so
we need to 6 H in product side also, so we need to put 3 H2O
C6H6(l) + O2(g) ----> 3 H2O(g) + 6 CO2(g)
C=6
C=6
H=6
H=6
O=2
O = 15
Now we need to balance the O – In the reactant there are 2 O and in the product there are 15 O,
so we need to 15 in reactant side also, so we need to put 15/2 O2
C6H6(l) + 15/2 O2(g) ----> 3 H2O(g) + 6 CO2(g)
C=6
C=6
H=6
H=6
O = 15
O = 15
Now we need whole number, so multiply by 2 to whole
2C6H6(l) + 15 O2(g) ----> 6 H2O(g) + 12 CO2(g)
C=6
C=6
H=6
H=6
O = 15
O = 15
So answer is – option c
Q 5 ) We are given, mass of Al = 82.79 g , mass of O2 = 117.65 g
First we need to calculate moles of each reactant
Moles of Al = 82.79 g / 26.98 g.mol-1 = 3.068 mol
Moles of O2 = 117.65 g / 32 g.mol-1 = 3.68 mol
Now we need to calculating the limiting reactant –
From the balanced equation
4 moles of Al = 2 moles of Al2O3
So, 3.068 moles of Al = ?
= 3.068 moles of Al * 2 moles of Al2O3 / 4 moles of Al
= 1.53 moles of Al2O3
From the balanced equation
3 moles of O2 = 2 moles of Al2O3
So, 3.68 moles of Al = ?
= 3.68 moles of Al * 2 moles of Al2O3 / 3 moles of O2
= 2.45 moles of Al2O3
So moles of Al2O3 is lowest from the moles of Al, so Al is limiting reactant.
Moles of Al2O3 = 1.53 moles
Mass of Al2O3 = 1.53 moles * 101.96 g/mol
= 156 g
So, answer is b. 155.8 g
Q 6 ) we are given, mass of Cl2 = 61.3 g
Mass of PCl5 = 164.7 g
First we need to calculate moles of Cl2 = 61.3 g / 70.906 g.mol-1
= 0.864 moles
Now we need to calculate moles of PCl5
From the balanced equation –
1 moles of Cl2 = 1 moles of PCl5
So, 0.864 moles of Cl2 = ?
= 0.864 moles PCl5
So, theoretical yield of PCl5 = 0.864 mole * 208.2 g/mol
= 179.99 g of PCl5
Percent yield = 167.7 g / 179.99 g * 100 %
= 93.2 %
So answer is near to e) 91.5 %
57. A chemistry student added 1.25 g of potassium carbonate in 125 mL of a 0.100 M acid
solution. The reaction can be represented by the following equation:
2HC6H7O7(aq) + CO32-(aq) → CO2(g) + H2O(l) + 2C6H7O7-(aq)
How many mole of acid is present?
How many mole of potassium carbonate is present?
What is the maximum volume (in mL) of the gaseous product being produced at 1 atm and
273.15K?
What is the % yield if only 77.5 mL of the gaseous product were collected?
ANSWER: mole of acid= Molarity x Volume= 0.1M x 0.125L= 0.0125 mol
mole potassium carbonate= mass/MW= 1.25g/138g/mol= 9.06 x 10-3 mol
Maximum volume being produced:
First find the limiting reactant
2 mole acid ----------- 1 mole potassium carbonate
0.0125 mole of acid ------- x= 6.25 x 10-3 mole of potassium carbonate. We have 9.06 x 10-3 mol
so, it is in excess, the limiting reactant is the acid.
2 mole acid -------- 1 mole CO2
0.0125 mole acid ---------x =6.25 x 10-3 mole of CO2
PV=nRT ----> V= 6.25 x 10-3 x 0.082L.atm/mol.K x 273.15K/1atm= 0.140L ---> 140mL
%yield= obtained amount/theoretical amount x 100= 77.5mL/ 140mL x 100= 55.4%
58.
ANSWER:
59. Calculate DeltaH, Deltas, and DeltaG for the following reaction at 25C (298.15K): N2O4(g)
--> 2N2O(g). Using a table, I've calculated (correctly according to the Mastering Chemistry
Gods) enthalpy at 57.2 (kJ) and entropy at 176 J/mol K. DeltaG by my 300 attempts is 4752, or
4.8 in kJ with 2 sig figs. Its wrong. So is 4.7. I've tried subbing in 298 instead of 298.15 (which
this stupid program prefers as the equivalent of 25, apparently). What am I doing wrong?
ANSWER: N2O4(g) --> 2N2O(g)
dH = 2*+9.2 - +33.2 = -14.8 kJ/mol
dS = 2*304.2 - 240.0 = 368.4 J/molK
dG = dH - TdS = -14800 -298*368.4 = -124583.2 J/mol = -124.5 kJ/mol
You are probably not including 2 mol of N2O
60. A 0.8870-g sample of a mixture of NaCl and KCl is dissolved inwater, and the solution is
then treated with an excess ofAgNO3 to yeild 1.913 g ofAgCl. Calculate the percent by mass of
each compound in themixture.
I think i have to come up with multiple algebra equations tosubstitute to solve but i can't figure it
out.
Molar mass
AgCl=143.35
NaCl=58.44
KCl=74.55
*It is number 102,chapter 4 in Raymond Chang's Fifthedition of General Chemistry
ANSWER: First let us make some observations:
The mass of AgCl is given, therefore the moles of AgCl and Cl-can be determined
1.913 g * (1 mol/143.35g) = .01334 moles of AgCl =.01334 moles of ClNow convert the Cl- into grams to find the remaining massof "K and Na"
.01334 moles of Cl * (35.453 g/ 1 mol) = .4729 gramsof ClNow, notice that both NaCl and KCl can be generalized as XCl(X has a charge of +1)
XCl exists in a 1:1 ratio similiar to AgCl, so .01334moles of X
Now, subtract this mass of chlorine from the originalmass:
.8870g -.4729g = .4141 grams of X
Find the molar mass of X (we use the hypothetical data, thegrams and moles found above)
.4141 g of X / .01334 moles of X = 31.04 g/mol(aka molar mass) of X
Now let
Na (molar mass = 22.98) be represented by Y (variable)
K (molar mass = 107.87) be represented by 1-Y
Y(22.98) + (1-Y)(107.87) = 31.04
-84.89Y = -76.83
Y = .9051
1-Y = .0949
Now.... multiply the remaining mass .4141 grams by theratios
So,
.9051 *.4141g = .3748 g of Na
.0949 *.4141g = .0393 g of K
Recall,
.4729 g of Cl
Total
= .887 g
-----------------------------------------------------------------------------------------Mass percent is (hopefully you can finish this step>.>
Mass of element
----------------- X 100 =
Mass of total
61. A demonstration many chemistry teachers like to perform in class isto combine aqueous lead
nitrate solution with aqueous potassiumiodide solution.
2 KI(aq) +Pb(NO3)2(aq) 2KNO3(aq) + PbI2(s)
Both reactants are colorless when the solutions are freshlyprepared, but the solid product is
bright yellow, so the processdemonstrates a double displacement very effectively.
Suppose a solution containing 1.15 gof KI is combined with a solution containing 2.56 g of
Pb(NO3)2. Whatmass of PbI2 would result?
ANSWER: The balanced equation says that 2moles of KI reacts with 1 mole of Pb(NO3)2 togive 1 mole
of PbI2.
mass of 1 mole of KI = molecular mass of KI = 1 * 39 + 1 * 127 =166g
=> 1 g of KI contains (1/166) moles
=> 1.15 g of KI contains (1/166) * 1.15 ~ 0.00693moles.
mass of 1 mole of Pb(NO3)2 = molecular massof Pb(NO3)2 = 1* 207 + 2 * 14 + 6 * 16 = 331g
=> 1 g of Pb(NO3)2 contains (1/331)moles
=> 2.56 g of Pb(NO3)2 contains (1/331) *2.56 ~ 0.00773 moles
Since we need the ratio in terms of number of moles ofPb(NO3)2 : KI to be 1 : 2 = 1/2 , clearlyPb(NO3)2 is
the limiting agent in this case.Thus, the reaction that takes place is that 0.00773 moles ofPb(NO3)2 reacts
with 0.00693/2 ~ 0.00347moles of KI to give 0.00773 moles of PbI2.
mass of 1 mole of PbI2 = molecular mass ofPbI2 = 207 + 2 * 127 = 461 g
=> mass of 0.00773 moles of PbI2 = 0.00773 * 461~ 3.56353 g.
62. I'm struggling with my chemistry question here,
Have 15 ml of 0.1 m KCl in test tube, also this solutionfreezes at -3.00 C.
molar mass of Kcl is 74.55 g/mol.
I need to find m = moles solute/ kg solvent which is =(grams/molecular weight)solute / kg
solvent.
Doing by myself, I found my mass of solvent to be (0.015L) * (1000m/1L) * (1Kg/1000g) *
(7455g/mL) =118.825kg.
I'm stuck after this. Any help QQ?
ANSWER: I'm struggling with my chemistry question here,
Have 15 ml of 0.1 m KCl in test tube, also this solutionfreezes at -3.00 C.
molar mass of Kcl is 74.55 g/mol.
I need to find m = moles solute/ kg solvent which is =(grams/molecular weight)solute / kg
solvent.
Doing by myself, I found my mass of solvent to be (0.015L) * (1000m/1L) * (1Kg/1000g) *
(7455g/mL) =118.825kg.
I'm stuck after this. Any help QQ
63.
At 100 degrees C, Keq = 0.078 for thefollowing reaction:
SO2Cl2(g) <-> SO2(g) +Cl2(g)
In an equilibrium mixture of the three gases the partial pressuresof SO2Cl2 and SO2 are 0.136 atmand
0.072 atm, respectively. What is PCl2 in theequilibrium mixture?
I use PV=nRT right?
Could you explain it showing steps? I skipped basic chemistry and Idon't recall knowing how to change
atm to moles (or Molarity)
ANSWER: We Know that :
The given equation is :
SO2Cl2(g) <-> SO2(g) +Cl2(g)
We Know that :
Kp = Kc ( RT )Δng
= 0.078 ( 0.0821 x 373 )
= 2.3886
Kp = PSO2 x PCl2 / PSO2Cl2
2.3886 = 0.072 atm x PCl2 / 0.136 atm
PCl2 = 2.3886 *0.136 / 0.072
= 4.511 atm
64. HO2 is a highly reactivechemical species that plays a role in atmospheric chemistry. Therate
of the gas-phase reaction
HO2(g) + HO2(g) -->H2O2(g) + O2(g)
is second order in [HO2],with a rate constant at 25 °C of 1.4 × 10^9 L/(mol*s).Suppose some
HO2 with an initial concentration of 2.0 × 10^-8M could be confirmed at 25 °C. Calculate
theconcentration that would remain after 1.0 s, assuming no otherreactions take place.
ANSWER: For second order reactions, the Integrated Rate Law is 1/[A] =kt + (1/[A]0)
[A] = concentration
[A]0 = initial concentration
k = rate constant
t = time in seconds
You plug everything you know into the equation. 1/[A] = (1.4 × 10^9)(1 s) + (1/ (2.0 × 10^ -8))
You solve for [A] and get about 6.9 × 10^-10 youdivide this by two because there are two HO2
And you get 3.5 × 10^ -10 M
65.
ANSWER: a. Cu + 2Ag+ = Cu2+ + 2Ag
b. K = [Cu2+]2 / [Ag+]
c. Eocell = EoAg+/Ag – EoCu2+/Cu = 0.80 – 0.34 = 0.46 V
dGo = -nFEocell =
= - 2 x 96486 C x 0.46 V = -88.8 kJ
d.lnK = - dGo/(RT) = 88 767 J/mol / (8.314 J/mol.Kx298K) = 35.83
K = e35.83 = 3.64x1015
e.
K is very high, so the reaction can be considered quite complete. According to the equation
stoichiometry
[Cu2+]= 0.155 / 2 = 0.077 M
66. Nitrogen dioxide, a pollutant in the atmosphere, can combine with water to form nitric acid.
One of the possible reactions is shown below.
a) Calculate ΔG∘ for this reaction at 25 ∘C.
3NO2(g)+H2O(l)→2HNO3(aq)+NO(g)
b) Calculate Kp for this reaction at 25 ∘C.
ANSWER: 3NO2(g)+H2O(l)→2HNO3(aq)+NO(g) : ΔG =?
ΔG = ΔG0f products – ΔG0f reactants
ΔG = [(2x ΔG0f
HNO3(aq)
)+ ( ΔG0f NO(g) )]- [( 3 x ΔG0fNO2(g) )+ (ΔG0f H2O(l) )]
= [(2x(-79.91))+((86.57))] - [(3x(51.30))+ (-237.18)] kJ/mol
= +10.03 kJ
= 10030 J
We know that ΔG = -RT ln K
Where
R = gas constant = 8.314 J/(mol-K)
T = Temperature = 25 oC = 25+273 = 298 K
K = Equilibrium constant = ?
Plug the values we get
ln K = -ΔG / (RT)
= -10030 / ( 8.314x298)
= -4.05
K = e-4.05 = 0.017
67. Use the following values to answer the questions below:
Ksp of AgOH = 1.50 x 10-8
Ka of HC2H3O2 = 1.74 x 10-5
-Suppose that AgOH is reacted with acetic acid (HC2H3O2). Use the above equilibrium
constants and your knowledge of acid-base chemistry to calculate the equilibrium constant for
the following reaction:
AgOH (s) + HC2H3O2 (aq) ----------> Ag+1(aq) + C2H3O2-1(aq) + H2O
ANSWER: AgOH
HC2H3O2
Ag+ + OH- Ksp = 1.50 x 10-8
C2H3O2- + H+ Ka =
---------------------------------------------------------------------Adding these two equations we get
AgOH + HC2H3O2
x 10-13
Ag+ +C2H3O2- +OH-+ H+ K = Ksp x Ka = 1.50 x 10-8 x 1.74 x 10-5 = 2.61
since OH- + H+ = H2O
The final reaction will be
AgOH + HC2H3O2
Ag+ +C2H3O2- + H2O; K = 2.61 x 10-13
68. Imagine that you are in chemistry lab and need to make 1.00L of a solution with a pH of 2.40. You
have in front of you 100 mL of 7.00�10?2M HCl, 100 mL of 5.00�10?2M NaOH, and plenty of distilled
water. You start to add HCl to a beaker of water when someone asks you a question. When you return
to your dilution, you accidentally grab the wrong cylinder and add some NaOH. Once you realize your
error, you assess the situation. You have 80.0mL of HCl and 85.0mL of NaOH left in their original
containers.
ANSWER: H 2.40 means [H+] = 10^(-2.40) M
The final volume must be 1.00L. That means the final excess (strong) acid HCl must be 10^(-2.40) Mole
15.0ml NaOH was accidentally added. Hence the total HCl needed is:
(10^(-2.40) + 0.0150*0.0500) Mole
or: (10^(-2.40) + 0.0150*0.0500)/0.0700 L
with 0.0150*0.0500 mole HCl to neutralize NaOH added.
Hence this many liters of HCl is needed:
(10^(-2.40) + 0.0150*0.0500)/0.0700 - 0.0200
69. The rate constant of a chemical reaction increased from 0.100 S^-1 to 2.60 S^-1 upon raising the
temperature from 25.0 C to 43.0 C.
Calculate the value of ln (k2/k1) .Please Note this is the second time I have posted this question I need it
in ln values not Log.
What is the activation energy of the reaction?Please note this is the second time asking this question.
The first attempt was wrong according to Mastering Chemistry. Check my other questions for the person
work.
ANSWER: According to Arrehenius Equation , K = A e -Ea / RT
Where
K = rate constant
T = temterature
R = gas constant = 8.314 J / mol - K
Ea = activation energy
A = Frequency factor(constant)
rate constant , K = A e - Ea / RT
log K = log A - ( Ea / 2.303RT ) ---(1)
If we take rate constants at two different temperatures , then
log K = log A - ( Ea / 2.303RT ) --- (2)
&
log K' = log A - ( Ea / 2.303RT' ) ---- ( 3)
Eq ( 3 ) - Eq ( 2 ) gives
log ( K' / K ) = ( Ea / 2.303 R ) * [ ( 1/ T ) - ( 1 / T' ) ]
Ea = [ ( 2.303R * T * T' ) / ( T' - T )] * log ( K' / K )
Given K = 0.100 / s
K' = 2.60 / s
T = 25+273 = 298 K
T' = 43+273 = 316 K
Plug the values we get Ea = [ ( 2.303R * 298 * 316 ) / (316-298 )] * log ( 2.6/0.1 )
= 141737 J
= 141.737 KJ
ln ( K' / K ) = ln ( 2.6/0.1 )
= ln 26
= 3.258
70. Find the molarity of an unknown solution of phosphoric acid if 12.0 mL of .20 M sodium
hydroxide is required to completely titrate a 20.0 mL sample of unknown phosphoric acid.
Please answer very simply and thoroughly. I am returning toschoolfor a 2nd bachelor degree and
it has been years since I lasthad amath or a chemistry class. Thank you in advance for your help.I
amgoing to post 2 other questions from the same lab workbook ifyoumight be able to help.
ANSWER: here we know that the product of molarity and volume isconstant.
so we have,
m1v1 = m2v2
now,
m2 = 12x .20 / 3 x 20
= 0.04 M
so molarity of phosphoric acid is = 0.04 M
71. Of the five group 2 metal hydroxides, Be(OH)2, Mg(OH)2, Ca(OH)2, Sr(OH)2, and
Ba(OH)2, berillyium hydroxide and barium hydroxide are defined as the most toxic. The goal of
this experiment was to determine the periodic trend in the solubility of group 2 hydroxide
compounds. Explain how the experiment you did today is more aligned with the philosiphy of
green chemistry than a similar experiment that involbed the titrations of all five group 2 metal
hydroxides
ANSWER: solubility order :
Be(OH)2 < Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2
Ba(OH)2 is most soluble and Be (OH)2 is least soluble.
one important point in this experiment . if lattice energy is more solubility is less and if
hydreation energy is more solubility is also more . based on this concept we get expermental
result.
lattice energy = K Z1 Z2 / r^2
in 2 nd group hydroxides Z1 = cation charge = 2
Z2 = anion charge = 1
these are common for all hydroxides in 2nd group.
but radius (r) increases with increase in size
'r ' value order : Be < Mg < Ca < Sr < Ba
so lattice energy is less for Ba(OH)2 .
if less lattice energy means it can easily break down into constituent ions.
while salt easily converted into ions these ions are easily hydrated . that means more amoun of
hydration energy is released.
Ba(OH)2 can be easily titrated with acid form salt . based on the reaction we can determine
amount of Ba(OH)2 present in the given any sample. we can estimate toxicity also
Ba(OH)2 + 2HCl -------------------------> BaCl2 + 2H2O
72. a) What is the concentration of a solution prepared my mixing 35.0mL of 0.10M acetic acid
with 15.0mL of water?
b.) Calculate [H3O+],[C2H3O2-], and [HC2H3O2] in this solution.
c.) What is the pH of this solution?
I know that I am supposed to set up an ICE table. I'm just not sure exactly how to do so. Buffers
aren't necessarily my strong point in Chemistry.
ANSWER: a) C1V1 = C2V2
C1 = 0.1M
V1 = 35.0 mL
C2 = ?
V2 = 15 +35 = 50 mL
C2 = C1V1/V2
C2 = 0.1M(35mL)/50mL = 7E-3 M
b) Ka = 1.8E-5
CH3COOH + H2O
Ci
Crx
Cfinal
CH3COO- + H3O+
CH3COOH
7E-3
-x
7E-3-x
H2O
CH3COO-
H3O+
+x
x
+x
x
Ka = [H3O+][CH3COO- ]/[CH3COOH ] = x2/7E-3-x
if x <<<<7E-3 then 7E-3-x = 7E-3
Ka = x2/7E-3
x = (Ka*7E-3)1/2
x= 3.5 E-4
x = [H3O+] =[CH3COO- ] = .5 E-4 M
[H3COOH]= 7E-3
c) pH = -log[H3O+] = 3.45
73. On a given day, the atmospheric pressure was measured at 73.2 cm HG. An industrious
general chemistry student, experimenting with non-volatile fluids in open ended manometers,
recorded data for a manometer filled with mineral oil (d=.822 g/mL). The level of the mineral oil
in the arm connected to the gas was 24.7 cm Higher than the arm open to the atmosphere. Given
that the density of mercury is 13.6 g/mL, determine pressure of the gas in the flask in atm.
ANSWER: Solution: Here, one end of manometer is open to gas-filled container and another end
is open to atmosphere. As, the level of the mineral oil in the arm connected to the gas is 24.7 cm
higher than the arm open to the atmosphere,
P1 + h = P0 or, P1 = P0 � h = 73.2 (cm Hg) � 24.7 (cm Hg) = 48.5 cm Hg = 0.638 atm.
So, the densities of the fluids are not required for the solution here.
74.
ANSWER: . the density is directly proportional to molar mass.
So, the molar mass of
CH4 = 16 g/mol
C2H4 = 28 g/mol
Kr = 84 g/mol
Ne = 20.17 g/mol
So, CH4 has the smallest density at STP.
Option A is correct
2.
According to ideal gas equation, Volume is directly proportional to moles of a gas.
Moles:
Ne = 10.0 g / 20.17 g/mol = 0.495 mol
Ar = 10.0 g / 40 g/mol = 0.25 mol
Kr = 10.0 g / 84 g/mol = 0.12 mol
Rn = 10.0 g / 210.9 g/mol = 0.0474 mol
Of these, 0.0474 mol is the least moles. So, Rn has smallest volume at STP.
Option D is correct
3.
From the given chemical reaction, it is clear that,
2 moles of Al2O3 is produced from 3 mol of O2.
So, given 4.89 g of Al2O3.
Moles of Al2O3 = 4.89 g / 101.96g/mol = 0.0479 mol
So, 0.0479 mol Al2O3 is produced from (0.0479 x 3) / 2 = 0.072 mol.
1 mol of gas at STP occupies 22.4 L
0.072 mol of gas at STP occupies 1.6128 L = 1612.8 mL
So, option D is correct
4.
V1/V2 = T1/T2
V1 = ?
V2 = 36.0 L
T1 = 15.7 degree C = 288.7 K
T2 = 22.3 degree C= 295.3 K
V1/36.0 = 288.7 / 295.3
=> V1 = 35.2 L
75.
Hydrogen iodide decomposes slowly to H2 and I2 at 600 K . The reaction is second order in HI and the
rate constant is 9.7 x 10^-6 M^-1 S^-1. If the initial concentration of HI is 0.110 M.
Part A: What is its molarity after a reaction time of 7.00 days?
[HI]=______________ M
Part:B What is the time (in days) when the HI concentration reaches a value of 8.0�10^-2 M?
t=________________ days
ANSWER: The equation is :
2 HI -> H2 + I2
You know the reaction is second order in HI, then the rate can be expressed as :
(1) r = k*[HI]^2
With k the rate constant ( k = 9.7 x 10^-6 M^-1 S^-1 )
The reaction rate can be defined in [HI] as :
(2) r = -1/2 * d[HI]/dt
By combining (1) and (2), you get the first order ordinary differential equation :
-1/2 * d[HI]/dt = k*[HI]^2
Which can be rewritten as :
-d[HI]/[HI]^2 = 2*k*dt
You can then integrate :
?-d[HI]/[HI]^2 = 2*k*?dt
1/[HI] = 2*k*t + c
You can find c for t = 0, [HI] = 0.130 M
1/(0.130) = c
c = 7.7 M^-1
You equation in [HI] and t is :
1/[HI] = t*2*9.7x10^-6 +7.7
(3) 1/[HI] = t*1.94x10^-5 + 7.7
Part A : t = 7 days = 7 * 24 * 60 * 60 = 604800 s
You replace t by its value in (3) :
1/[HI] = 19.4
[HI] = 0.05 M
Part B : [HI] = 8.5�10?2 M
You replace t by its value in (3) :
1/(8.5�10?2) = t*1.94x10^-5 + 7.7
t*1.94x10^-5 = 4.06
76. (Ideal behavior) Consider the reaction of nitric oxide and oxygen to form nitrous oxide in a
closed container. The container initially held 1.39 g of oxygen and 2.432 g of nitric oxide at a
pressure of 1.256 atm and a temperature of 34.5oC. What is the pressure after the reaction if the
volume and the final temperature are uncharged. Give your answer to the correct number of
significant figures. The reaction to consider is 2 NO + O2 -----> 2 NO2
(We often overlook the real chemistry that will occur, what is probably the most important
complication we have overlooked?)
ANSWER: 2 NO + O2 -----> 2 NO2
No of moles of NO = 2.432/30 = 0.0811 mole
No of moles of O2 = 1.39/32 = 0.0434 mole
initial Total moles = 0.0811+0.0434 = 0.1245 mole
volume of container = 0.1245*0.0821*(34.5+273.15)/1.256
= 2.5 L
from equation 2 mole NO = 1 mole O2
limiting reagent is NO
No of moles of NO2 produced = 0.0811 mole
No of moles of O2 unreacted = 0.0434- 0.0405 = 0.0029 mole
total no of moles after the reaction = 0.0029+0.0811 = 0.084 mole
final pressure P = 0.084*0.0821*(34.5+273.15)/2.5 = 0.84 atm
77. This problem is #64, chapter 17, Chemistry 8th Ed., Zumdahl. Hydrogen sulfide can be
removed from natural gas by the reaction: 2H2S(g) + SO2(g) <=> 3S(s) + 2H2O(g). Calc ΔGo and K
(at 298K) for this reaction. Would this reaction be favored at a high or low temperature?
I found ΔGo = -90kJ/mol. I'm having trouble with calculating K and understanding how the
reaction would favor a hi or lo temp.
.
ANSWER: To calculate ΔGo
It is easiest to use the standard formation of ΔGo = ΣnΔG°f(products)−ΣmΔG°f(reactants)
I am assuming you have done so correctly.
Free energy change is related to the reaction quotient, in that ΔG= ΔG°+ RT lnQ
Note that at equilibrium Q = K and ΔG = 0.
Therefore 0 = ΔG°+ RTlnK
and K = e^(−ΔG°/RT).
Note that at Standard conditions T = 298 Kelvin and R = 8.314
To determine whether the reaction is favored at high/low temperature, we must remember how
ΔG°is related to temperature.
ΔG°T≈ΔH°298 K−T(ΔS°298 K)
In this case entropy notably decreases as 3 moles of gas becomes 2 moles of gas.Enthalpy, like
Gibbs free energy, can be determined using the standards of formation
ΣnΔH°f(products)−ΣmΔH°f(reactants).
If enthalpy decreases (exothermic) and entropy decreases, the reaction will be favorable at low
temperature.
If enthalpy increases (endothermic)and entropy decreases , the reaction will never be favorable at
any temperature.
If enthalpy increases (endothermic) and entropy increases, the reaction will be favorable at high
temperatures.
If enthalpy decreases(exothermic) and entropy increases, the reaction will be favorable at all
temperatures.
78. my teacher told me to add the equations but before doing that I have to multiply them for things to
cancel out... I am so confused. Its an online chemistry class, so hard to teach myself! thanks!
H2O2 can be prepared by the following reaction, H2(g) + O2(g) ? H2O2(l)
Calculate the ?H for 1 mole, using the following data:
2H2(g) + O2(g) ? 2H2O(l); ?H = -571.6kJ
H2O(l) + 1/2O2(g) ? H2O2(l); ?H = 98.0 Kj
ANSWER: The answer is: -187.8 kJ. See solution below:
(1) 2 H2(g) + O2(g) => 2 H2O(l); ΔH1 = -571.6 kJ
(2) H2O(l) + 1/2 O2(g) => H2O2(l); ΔH2 = 98.0 kJ
Add equation (1) + 2 x equation (2):
2 H2(g) + O2(g) + 2 H2O(l) + O2(g) => 2 H2O(l) + 2 H2O2(l)
Cancel common terms to get:
2 H2(g) + 2 O2(g) + O2(g) => 2 H2O2(l)
Divide by 2 to get the required reaction:
H2(g) + O2(g) => H2O2(l)
Add up and divide ΔH in the same way:
ΔH(reaction) = (ΔH1 + 2 x ΔH2)/2
= (-571.6 x 2 x 98.0)/2
= -187.8 kJ
79.
ANSWER: For a first order reaction rate constant , k = ( 2.303 /t )* log ( a / (a-x))
Where
a = initial concentration = 100 ( say)
a-x = concentration left after time t = 63.16
t = time = 23 years
Rate constant = ?
Plug the values we get k = ( 2.303 / 23 ) * log(100 / 63.16)
= 0.01998 year^-1
For a first order reaction ,
Half life , T = 0.693 / k
= 34.63 years
80.
The enthalpy of vaporization of ethanol is 38.68 kJ/mol at its boiling point (78oC). Calculate the value of
?Ssurr when 1.00 mole of ethanol is vaporized at 78oC and 1.00 atm.
A. 1.10e2 J/K mol
B. 4.96e2 J/K mol
C. –1.10e2 J/K mol
D. –4.96e2 J/K mol
E. 0
ANSWER: We know that :
dSsurrounding = - ΔHvaporization / T (K)
ΔHvap. = 38.68 kJ/mol
= 38680 J/mol
Tboiling = 780C
= ( 78 + 273 ) K
= 351 K
dSsurr. = - 38680 J/mol / 351 K
= - 110.19 J/mol-K
= -1.10 x 10-2 J/mol-K
option c is correct.
81.
ANSWER: We know that rate of effusion , r α 1 / √M
Also r = V / t
From the above two equations t α √M
t / t' = √( M / M')
Where
M = molar mass of gas with emperical formula CH = ?
M' = molar mass of Ne = 20.18 g/mol
t = time of diffusion of CH = 39.5 min
t' = time of effusion of Ne = 20.0 min
Plug the values we get M / M' = ( t / t' ) 2
M = M' x ( t / t' ) 2
= 20.18 x ( 39.5 / 20.0 ) 2
= 78.7 g/mol
Emperical mass of CH is = 12 + 1 = 13 g/mol
So integer , n = molar mass / emperical mass
= 78.7 / 13
= 6.05
~6
So molecular formular is 6(CH) = C6H6
-------------------------------------------------------------From ideal gas law PV = nRT
--> PV / T = P'V' / T'
Where
P = initila pressure
P' = final pressure = 3P
T = initial temperature
T' = final temperature = T/2
V = initial volume = 12 L
V' = final volume = ?
Plug the values we get V' = PVT' / P'T
= ( P x 12 x (T/2) ) / ( 3P x T)
=2L
82. fluoride, (HF) at a temperature above the boiling point of water, so you are dealing with water vapor
and not just a pure liquid:
SiO2(s) + 4HF(g) -- SiF4(g) + 2H2O(g)
Predict the effect and why (explain) on [SiF4] when
a) H2O(g) is removed
b) Some liquid water is added
c) HF is removed
d) Some SiO2 is removed
ANSWER: The equilibrium changes by the following changes as SiO2(s) + 4HF(g) <--> SiF4(g) + 2H2O(g)
a) If H2O is removed on the product side the equilibrium shifts towards right.This indicates that more
SiF4 is formed.
b) No effect on the reaction even we add some liquid water. So reaction is not changed.
c) If HF is removed equilibrium shifts to wards left to form more Hf not [SiF4] hence the production of
[SiF4] will be decreases.
d) The removing of SiO2 not effects the reaction because the equilibrium doen't effects by changing the
solid or liquid concentrations.
83. An aqueous solution contains glycine. It doesn't ionize in water calculate molality if solution
freezes at-.8 degree C???
show work and i know you use Delta Tf= Kf times molality but i cant find the Kf value and need
to know where it is that i find these values in the book. My book is chemistry 9th edition by
raymond chang. Its grey with 3 pics on the front. I cannot find it anywhere in this book and tried
googling the Kf value but wasn't successful
ANSWER: We know that ΔT f = Kf * m
Where
ΔT f = depression in freezing point
= Freezing point of water - freezing point of solution
= 0 oC - ( -0.8 oC )
= 0.8 oC
K f = depression in freezing constant for water = 1.86 oC / m
( These are the standard values )
m = molality of the solution
Plug the values we get m = 0.8 / 1.86
= 0.4301 m
84. A chemistry student with a mass of 75 kg is riding a steadily moving Ferris wheel. When she
is at the top of the Ferris wheel, the normal force from the seat on to her body has a magnitude of
557 N.
a) What is the magnitude of the normal force on her body when she is at the bottom of the Ferris
wheel's arc?
913 N was the right answer
b) What would the normal force be on the student at the top of the wheel if the wheel's velocity
were doubled?
ANSWER: If the wheels velocity were doubled mv^2 / R increases by a factor of four, i.e
Weight of the student = 75 kg * 9.81 = 735.75 N - Normal Force ( 556 N )
= 179.75 N * 4 ( coz, it increases by a factor of 4 )
= 719 N
There fore Normal force = 719 N - 735.75 N = 16.75 N , is this correct!!!!!!!
:confused:
Your numerical answer is probably correct, but your "stream of consciousness" approach to solving
problems is going to get you in trouble when solving more compex problems. Do yourself a big favor and
resolve to never put an equal sign between two things that are not equal. As a case in point
719 N - 735.75 N = 16.75 N
Well, no. This statement is false. In truth
719 N - 735.75 N = -16.75 N
A negative normal force is in the realm of possibility in this problem, as long as there is a bar or belt to
hold in the passengers. How do you know the answer is not negative?
Also,
Weight of the student = 75 kg * 9.81 = 735.75 N - Normal Force ( 556 N )
You just wrote a contradiction. Ignoring for the moment the units that you left out, you said
W = 75 * 9.81 = 179.75
Since when was 75*9.81 = 179.75?
Try rewriting your solution in such a way that every statement you make is true. If you say A = B, then
make sure A in fact does equal B.
85. A chemistry student prepares lead(II) iodide from 1.35 g of lead (II) nitrate and excess
aqueous potassium iodide. If the student collected 1.72 g of PbI2 and the theoretical yield is 1.88
g, what is the percent yield? A 1.95 g sample of potassium bicarbonate is decomposed by
heating. If the resulting potassium carbonate weighs 1.54 g and the calculated yield is 1.35 g,
what is the percent yield? can anyone help me with either of these I'm very lost D:
ANSWER: Pb(NO3)2 + 2 KI = PbI2 + 2 KNO2
moles PbI2 = 1.72/461.05
theoretical amount PbI2 = 1.88/ 461.05
% yield = 1.72/ 1.88 = 91.489 %
86. An sample of unknown compound consiting of some amount of Ligand ethyldiamine
molecules NiSO4�6H2O 0.1566 g of this substance is dissolved in 10 ml H2O and
titrated with 27.4 ml of 0.103M HCl to endpoint.
H2N-CH2-CH2-NH2(aq) + 2 HCl(aq) <--> [H3N-CH2-CH2-NH3]^2+(aq) + 2Cl^-(aq)
what is the mass of ethyldiamine in the compound? and what is the percent ethyldiamine in this
sample?
ANSWER: H2N-CH2-CH2-NH2(aq) + 2 HCl(aq) <=> [H3N-CH2-CH2-NH3]^2+(aq) + 2Cl^-(aq)
Moles of HCl = volume x concentration of HCl
= 27.4/1000 x 0.103 = 0.0028222 mol
Moles of ethyldiamine = 1/2 x moles of HCl
= 1/2 x 0.0028222 = 0.0014111 mol
Mass of ethyldiamine = moles x molar mass of ethyldiamine
= 0.0014111 x 60.10
= 0.08481 g = 0.0848 g
Mass percent of ethyldiamine = mass of ethyldiamine/mass of sample x 100%
= 0.08481/0.1566 x 100%
= 54.16% = 54.2%
87. Im trying to do a lab report for my chemistry class, I cannot figure out how to find delta H
and delta S.
The delta G for the first trial is 26921.1 J at 295.0 K and 40992.9 J at 362.1 K.
I know that you use delta G = delta H - Tdelta S and make an equation for both, but im lost from
there.
40992.9 = delta H - (362.1)deltaS
26921.1 = delta H - (295) delta S
ANSWER: 40992.9 = delta H - (362.1)deltaS
26921.1 = delta H - (295) delta S
substract equation 2 from equation 1
14071.8 = -67.1 delta S
so deltaS= -14071.8 /67.1
deltaS= -209.714 J/K
substitute this value in equation 1
40992.9 = deltaH- 362.1 x -209.714
40992.9= deltaH+ 75937.388
detaH= 40992.9 - 75937.388
88. A solution contains Ag+ and Hg2+ ions. The addition of 0.100 L of 1.41M NaI solution is
just enough to precipitate all the ions as AgI and HgI2. The total mass of the precipitate is 32.6g .
Find the mass of AgI in the precipitate.
Express your answer to two significant figures and include the appropriate units.
ANSWER: Ag+(aq) + Hg2+(aq) + 3I-(aq) = AgI(s) + HgI2(s)
moles I- added = (Molarity I-)(L of I-) = (1.41)(0.100) = 0.141 moles I- added
The balanced equation tells us that Hg2+ consumes twice as much I- in the reaction as
does Ag+, since the formula HgI2 contains 2 I- atoms while the formula AgI contains
only 1 I- atom. So if Ag+ consumes x moles of I-, then Hg2+ consumes 2x moles of I-.
The total moles of I- consumed = 3x = 0.122 moles I-.
3x = 0.141; x = 0.141 / 3 = 0.0705 moles I0.0705 moles I- x (126.9 g I- / 1 mole I-) = 8.95 g IThe formula AgI tells us that the mole ratio of Ag+ to I- is 1:1.
0.0705 moles I- x (1 mole Ag+ / 1 mole I-) = 0.0705 moles Ag+
0.0705 moles Ag+ x (107.9 g Ag / 1 mole Ag) = 7.61 g Ag+
So the mass of AgI = mass of Ag+ + mass of I- = 7.61 + 8.95 = 16.56 g AgI
89. A chemistry professor has a cup of coffee containing 50.0mL of room temperature coffee at
25.0C. The professor also has a new pot of hot coffee at temperature of 96.0C. What volume of
hot coffee will the professor need to add to his cold coffee to reach the ideal drinking
temperature of 82.0C? Assume that no heat is lost to the coffee cup or to the environment. Also
assume that coffee has the same density (1.0g/mL) and heat capacity (4.184 J/gC) as water.
ANSWER: let m be the mass of water
we have
50 x 4.184 x (82 - 25) = m x 4.184 x (96 - 25)
m = 203.67 g
volume = 203.67 g/mL
Therefore, the answe ris 203.67
90. A large gas flask is found to weigh 143.567 g. it is then filled with gas to a pressure of 735 torr at 31 C and found to weigh
1067.9 g (the density of water is 31 C is 0.997 g/mL). Assuming that the ideal gas equation applies, calculate the molar mass of
the unknown.
ANSWER: P = 735 torr = 0.967 atm
m = 1067.9-143.567 gm = 924.333gm
d = 0.997 g/ml
V = 927.1 ml = 0.9271ml
T = 31 C = 304 K
PV=m/M RT
R = 0.082
M = mRT/PV = 25701.75 gm
91. The calcium carbonate- calcium bicarbonate equilibrium is particularly important for the
field of aquatic and environmental chemistry. CaCO3(s) + CO2 + H2O(l) = Ca(HCO3)2 (aq) a)-
Demonstrate the equilibrium constant equation for the reaction given below. b)- An experiment
in the laboratory determined that a reaction that a reaction mixture containing 11.935 g of
CaCO3, 3.86x10^-2 mol/L of CO2, 250 L of H2O and 1.82x10^-7 mol/L of Ca(HCO3)2 had
reached equilibrium . Using this information, calculate the equilibrium constant for the calcium
carbonate- calcium bicarbonate equilibrium.
ANSWER: K=[Ca(HCO3)2]/[CO2] (since for liquid and solids the concentration is taken to be 1)
so Keq=3.86*10^-2/1.82*10^-7
=2.12*10^5
92. The Ideal Gas Law is written as follows: PV = NK_BT. (The variable N is the number of
particles,
not the number of moles, as you've seen in chemistry. The variable K_B is the Boltzmann
constant,
where K_B = 1.38*10^-23 m2kg/s^2K in SI units.) Clearly state the meaning and the units of
variables P,
V , and T. Check that each side of PV = NK_BT has units of energy (joules in the SI system).
ANSWER: Check that each side of PV = NK_BT has units of energy (joules in the SI system):
W = Fd
Joul = N.m = (kg.m/s^2).m = m^2kg/s^2
[k_B] [T] = (m2kg/s^2K)(K) = m2kg/s^2 = m.kg.(m/s^2)
Therefore NK_BT has units of energy.
[P][V] = (N/m2) * (m3) = N.m = J
Therefore PV has units of energy.
93. Hydrogenation of carbon double bonds C=C (i.e. adding a hydrogen on each carbon atom
and thus reducing the double bond into a single bond: H-C-C-H) is a very common process in
organic and polymer chemistry. Using the data from the following table, calculate the ?Hreaction
when ethylene C2H4 is reduced into ethane according to the reaction below: C2H4(g) + H2(g) -> C2H6(g) Bond Bond enthalpy (kJ/mol) C=C 620 C-C 347 H-H 436 C-H 414 The ?Hreaction is
1.----- kJ/mol
ANSWER: given reaction
C2H4(g) + H2(g) --> C2H6(g)
given data : Bond enthalpy (kJ/mol) C=C 620 C-C 347 H-H 436 C-H 414
therefore let, dH rxn = dBond energies reactants - products
dH rxn = [H-H @463 & C=C @ 620 & 4 C-H @ 414 each] - [C-C @ 347 & 6 C-H @ 414 each]
dH rxn = [463+620+(4*414)] - [347+(6*414)]
dH rxn = [2,739] - [ 2831] = -91 Kj
94. When 1.40g of magnesium metal is allowed to react with 200ml of 6.00M aqueous HCl , the
temperature rises from 25.0 C to 42.3C . Calculate delta H in kilojoules for the reaction,
assuming that the heat capacity of the calorimeter is 776J/C , that the specific heat of the final
solution is the same as that of water [(4.18 J/(gxC) ], and that the density of the solution is
1.00g/mL . Chemistry
ANSWER: Mg + 2 HCl => MgCl2 + H2. Moles of Mg = mass/molar mass of Mg = 1.40/24.305 =
0.05760 mol. Heat released by reaction = heat absorbed by solution + heat absorbed by
calorimeter = mass x specific heat x temperature change of solution + heat capacty x temperature
change of calorimeter = (200 + 1.40) x 4.18 x (42.3 - 25.0) + 776 x (42.3 - 25.0) = 27988.84 J.
Delta H = -heat released/moles of Mg = -27988.84/0.05760 = -485917 J/mol = -486 kJ/mol.
95. A chemistry student with a mass of 75 kg is riding a steadily moving Ferris wheel. When she
is at the top of the Ferris wheel, the normal force from the seat on to her body has a magnitude of
557 N.
a) What is the magnitude of the normal force on her body when she is at the bottom of the Ferris
wheel's arc?
b) What would the normal force be on the student at the top of the wheel if the wheel's velocity
were doubled?
ANSWER: (a)
mg - Ntop= Fc , where Fc= centripetal force required
75x9.8 - 557 = Fc
∴ Fc= 178 N
now when she is at bottom,
Nbottom - mg = Fc
∴ Nbottom - 75x9.8 N= 178N
∴ Nbottom =178 + 75x9.8 = 913 N (ans.)
(b) Fc = mv2/r
when v doubles,Fc will increase to 4 times its initial value .
therefore in the second case,new Fc = 4x557 N = 2228 N
∴ Now, mg - Ntop= Fc
75x9.8 - Ntop= 2228
∴ Ntop = 2228 + 75x9.8 = 2963 N in the upward direction (ans.)
96.
ANSWER:
176.12 g/mol = molar mass of ascorbic acid
mole of acid = 00.302/176.12
number of equivalent = 0.302/176.12 *2
we have to balance the number of equivalent of acid and base
so,
0.302/176.12*2 = 0.12*V
V = 0.0285L
97. Juan makes a measurement in a chemistry laboratory and records the result in his lab report.
The standard deviation of students' lab measurements is ? = 4 milligrams. Juan repeats the
measurement 3 times and records the mean x of his 3 measurements.
(a) What is the standard deviation of Juan's mean result? (That is, if Juan kept on making 3
measurements and averaging them, what would be the standard deviation of all his x's?) (Use 3
decimal places.)
1 Incorrect: Your answer is incorrect.
ANSWER: In short, what your are talking about is commonly called the "standard error of the mean."
When you take one measurement the standard deviation is for the population. If you take x number of
measurements, you can calculate the standard deviation of the sampling distribution of size x. The
formula for this is the standard deviation of the population divided by the square root of the number of
measurements.
Basically any one given measurement is likely to be off to a certain extent from the real value. With
more measurements, some measurements will be smaller and others will be larger than the true value,
and these differences will tend to cancel each other out, making the standard deviation of the sample of
measures smaller than if you went with one measurement.
As far as reducing the standard deviation to 5:
5=10/sqrt(x)
Rearranging terms we get:
sqrt(x)=10/5=2
(sqrt(x))^2=2^2
x=4
98.
Determine the thickness of a sheet of aluminum foil
An Introductory Chemistry student is given a square of aluminum foil, and is told the density of
aluminum is 2.70 g/cm3
. The student�s task is to determine the thickness of the foil.
The student measures the length and width of the square of foil with a ruler, weighs it on a
balance, and
records the following:
Length = 15.4 cm
Width = 15.3 cm
Mass = 1.650 g
Calculate area of foil: ______________________
Calculate volume of foil: ______________________
Calculate thickness of foil: ______________________
ANSWER: Area of foil = length x width
= 15.4 x 15.3 = 235.62 cm2 (or approximately 236 cm2)
Volume of foil = mass/density
= 1.650/2.70 = 0.6111 cm3 (or approximately 0.611 cm3)
Thickness of foil = volume/area
= 0.6111/235.62
= 0.002594 cm = 0.00259 cm = 2.59 x 10^(-3) cm
99.
ANSWER: (a) DHvap = 60.0 kJ/mol = 60000 J/mol
DSvap = 100.0 J/mol.K
q(surr) = -DHvap = -60000 J/mol
Temperature T = 27 deg C = 300.15 K
DS(surr) = q(surr)/T
= -60000/300.15
= -200 J/mol.K = -0.200 kJ/mol.K
(b) At the boiling point:
DGvap = DHvap - Tb x DSvap = 0
Boiling point Tb = DHvap/DSvap
= 60000/100.0
= 600 K
100. K2Cr2O7 + HCl => CrCl3 + KCl + Cl2 + H2O
FeCO3 + H2SO4 => Fe2(SO4)3 + S + CO2 + H2O
Al + HNO3 => Al(NO3)3 + N2 + H2O
K2Cr2O7 + SO2 + H2SO4 => Cr2(SO4)3 + K2SO4 + H2O
Mg + H2SO4 => MgSO4 + H2S + H2O
ANSWER: balance first
14HCl + K2Cr2O7 -> 2KCl + 2CrCl3 + 7H2O + 3Cl2
Oxidation reaction
6Cl- -> 3Cl2
1. redox reaction
2Cr+6 -> 2Cr+3
half reaction
6Cl- + 2Cr+6-> 3Cl2 -> 2Cr+3