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Chemistry 534
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Name_________________
2010 Pretest (wasTest 2.1v2005-06)
1.
One word answer: Potential? Or Kinetic?
a.
%
Feeling Exo? Or Endo?__
(energy)
(3 marks)
Gasoline’s energy while it’s sitting in your gas tank.
Mostly potential. Will release energy when burnt.
b.
All intramolecular and intermolecular bonds within and between sugars.
Potential energy
c.
A hurricane sweeping across Florida.
Kinetic energy= energy of motion
2.
TRUE? or FALSE?
(4 marks)
a. An endothermic reaction causes its environment to experience a drop in temperature.
True. It will steal energy from the environment, causing the temperature of its
surroundings to drop.
b. If a compound contains 300 kJ of energy and it breaks up into two elements each
containing 160 kJ, then the reaction should have released 20 kJ of heat into the
environment.
FALSE (160 + 160) – 300 = + 20 kJ = endothermic, so it needs to absorb or steal 20 kJ
from its environment, not release.
c. Enthalpy is either the heat content of a substance or its sum of potential and kinetic
energies.
TRUE
d. Each C3H8 molecule forms 10 intramolecular bonds.
TRUE. There are two ways of getting the number of intramolecular bonds. Either draw a
structural formula or use the following formula:
Where C = number of carbons in formula; N = # of nitrogens; O = # of oxygens;
H = # of hydrogens or halogens.
H
H
C
C
H
2.
H
H
C
H
H
H
When phosphorus is burned in an excess of oxygen, tetraphosphorus decoxide(P4O10) is
formed as shown:
4 P(s) + 5 O2(g)
ΔHf = ?
P4O10(s)
Use the equations below to calculate the ΔH of formation, ΔHf, of tetra phosphorus
decoxide.
(4 marks)
ΔH = -587 kJ
Equation(A)
2 PCl3(l) + O2(g)
2 POCl3(l)
Equation(B)
P4O10(s) + 6 PCl5(s)
Equation(C)
2 P(s) + 3 Cl2(g)
2 PCl3(l)
ΔH = -686 kJ
Equation(D)
2 P(s) + 5 Cl2(g)
2 PCl5(s)
ΔH = -892 kJ
10 POCl3(l)
Reverse (B)
10 POCl3(l)
5 X Equation(A)
10 PCl3(l) + 5O2(g)
P4O10(s) + 6 PCl5(s)
To get rid of PCl3 ,5 X (C) 10P(s) + 15 Cl2(g)
Reverse& 3X (D)
6 PCl5(s)
4 P(s) + 5 O2(g)
ΔH = -419 kJ
10 POCl3(l)
ΔH = -587 kJ(5)
10 PCl3(l)
ΔH = -686 kJ(5)
6P(s) + 15 Cl2(g)
P4O10(s)
ΔH = + 419 kJ
ΔH = +892 kJ(3)
ΔHf = -3270 kJ
3.
Find the molar heat of combustion, in kJ/mole, of magnesium(Mg) if a 1.00 g sample
warms one liter of water from 10.00 oC to 17.83 oC.
(4 marks)
Reactant = Mg
1.00/24.3g/mole = 0.0411522…moles Mg
Env = water
=1000.g(4.19J/(goC)(17.83-10.00)
= 32807.7 J
= -797 kJ/mole of Mg
4.
Redraw and then correctly place the following on the graph shown below: a burnt
chicken, an uncooked chicken, a well-cooked chicken, and an
undercooked chicken. Note “ progress of reaction” is on the graph’s xaxis:
(2 marks)
(4 marks)
Well cooked
undercooked
E(kJ)
Uncooked chicken
Burnt chicken
Reaction progress
5.
What temperature change will be experienced by a 250.0 mL mixture (final volume) if it
resulted from the neutralization of equal amounts of NaOH and HI?
Molar heat of neutralization of NaOH= -200.0 kJ/mole
Concentration of NaOH = 0.30 g/L
(4 marks)
Reactant = NaOH
Since equal amounts of acid and base were
used, volume of NaOH = 250.0 ml/2 = 125 ml
= 0.125 L
m = CV = 0.30 g/L(0.125 L) = 0.0375 g NaOH
Env = water
0.0375 g NaOH/(40.0 g/mole )= 9.375 X10-4
moles
9.375 X 10-4 moles*(-200.0 kJ/mole)
= -0.1875 kJ
We assume that solutions are aqueous and with
density of 1.00g/ml
=250.0g(4.19J/(goC)( )
= 0.18 oC
The reason the temp change is so small is
because 0.30 g/L is a very dilute solution. In
the lab we used 2.0g/0.100L = 20g/L, which is
over 60 times stronger.
6.
The dissolution of NaOH(s) in water is represented by the following chemical equation:
Equation(A) NaOH(s)
Na+(aq) + OH-(aq) + 42 kJ
The neutralization of NaOH(aq) by HCl(aq) is represented by the following chemical
equation :
Equation(B) Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq)
Na+(aq) + Cl-(aq) + H2O(l)
H = -63 kJ
Find
H for the following reaction:
NaOH(s) + H+(aq) + Cl-(aq) Na+(aq) + Cl-(aq) + H2O(l)
(4 marks)
Keep Equation(A)
NaOH(s)
Na+(aq) + OH-(aq) + 42 kJ
Keep Equation(B)
Equation(B) Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq)
NaOH(s) + H+(aq) + Cl-(aq)
7.
Na+(aq) + Cl-(aq) + H2O(l)
H = -63 kJ
Na+(aq) + Cl-(aq) + H2O(l)
The Café Entropy makes cappuccino.
Cappuccino is a mixture of coffee and milk.
The Café Entropy has determined that the best temperature for cappuccino is
45.5 C.
The initial temperature of hot coffee without milk is 70.5 C.
What volume of milk, at 4.0 C, must be added to 160.0 mL of hot coffee
in order to obtain the desired temperature of 45.5 C?
(note: Assume coffee and milk have the same density and specific heat capacity as
water.)
(4 marks)
Notice that 4.19 cancels
4000 = 41.5m
m = 96 g of milk = 96 ml
8.
Hydrogen can be prepared by the action of hydrochloric acid on zinc. What mass
of HCl must react to produce 50.0 litres of hydrogen measured at 300.0K and at a pressure of
101 kPa?
(4 marks)
Equation : Zn(s) + 2 HCl(aq)
ZnCl2(aq) + H2(g)
= 2.025…moles H2
2.025…moles H2
9.
Four cylinders filled with gas are at the same temperature. In which of the following will
the gas exert the most pressure?
(2 marks)
A)
B)
C)
D)
A 2.0 L cylinder containing 4.0 g of H2.
A 2.0 L cylinder containing 4.0 g of He.
A 1.0 L cylinder containing 16.0 g of O2
None. At constant temperature, pressure is constant.
Since P is proportional to density:
, the one with the highest density will have the
highest pressure
(a)
(b)
(c)
(d)
(4.0/2.0g/mole)/2.0 L = 1.0 mole/L =Answer
(4.0/4.0g/mole)/2.0 L = 0.50 mole/L
(16/32.0g/mole)/1.0 L = 0.50 mole/L
nonsense