Download Grade 11 review answers

Grade 11 Chemistry Review Answers
1) a) If 35.0 g of Iron reacts with Oxygen to produce a substance with a total mass of 50.0 g, what
mass of Oxygen must have reacted?
50.0g - 35.0g = 15.0 g (Law of conservation of mass)
b) In another experiment, 58.4 g of Iron reacts with Oxygen to produce a substance with a total mass
of 75.5 g. Is this substance the same as the substance in question 3a? Explain.
35.0g / 50.0g = 0.700 mass fraction
58.4g / 75.5g = 0.774 mass fraction
Since the mass fractions (percent compositions if we were to multiply by 100%) are different, the substances are different
(Law of definite proportions).
2) Element X has two isotopes: 297 and 301. 25% of the atoms in random samples are of the lighter
isotope. Calculate the average atomic mass for this element.
We will assume that the atomic mass of each isotope is about equal to its mass number.
Average atomic mass for element X: AR = (0.25 x 297u) + (0.75 x 301u) = 3.0 x 102u
3) Construct Bohr diagrams for the following atoms: a) N, b) Ca, c) O2-, d) Mg2+, e) S
4) Define the following terms:
a) first ionization energy
The energy needed to remove the first electron from a neutral atom.
b) electron affinity
The energy released when an electron is added to a neutral atom.
c) electronegativity
The degree to which electrons in a covalent bond are attracted to an atom.
d) atomic radius
One half of the distance between two covalently bonded atoms of the same element.
5) Distinguish between the following types of chemical bonds:
a) ionic bond
The electrostatic attraction between oppositely charged ions.
b) covalent bond
The mutual attraction of two nuclei for a shared pair of electrons.
c) polar covalent bond
The unequal mutual attraction of two nuclei for a shared pair of electrons.
d) co-ordinate covalent bond The mutual attraction of two nuclei for a shared pair of electrons where originally a lone
pair on one atom.
6) Consider the compound Sodium acetate; classify each bond as
C-C and C-H
a) purely covalent
b) polar covalent
C-O and C=O
c) ionic
O- and Na+
7) i) Use Lewis Dot diagrams to illustrate the bonding between atoms of the following elements.
Indicate the type of bond in each case (purely covalent, polar covalent (indicate partial positive and
negative poles), and ionic (indicate the positive and negative ion).
ii) Give the correct IUPAC name and chemical formula of the resulting compound.
a) Sulfur and Sodium
b) Carbon and Chlorine
c) Calcium and Nitrogen
d) Aluminum and Sulfur
e) Hydrogen and Nitrogen d) Nitrogen and itself
a) I: Na+ S2- b) PC: Cl- C+
c) I: Ca2+ N3- d) I: Al3+ S2e) PC: N- H+ e) C
8) Give the correct IUPAC name for the following compounds:
a) PbO2 b) N2O4 c) CuNO3 d) CCl4 e) Sn3(PO5)4 f) H2CO2(aq) g) MgSO4A7H2O
a) Lead (IV) oxide
d) Carbon tetrachloride
g) Magnesium sulfate heptahydrate
b) dinitrogen tetroxide
e) Tin (IV) phosphate
c) Copper (I) nitrate
f) dihydrogen Carbon dioxide (propanone)
9) Write the correct chemical formula for the following compounds
a) Mercury (II) nitride
b) Calcium nitrite
c) Iron (III) acetate
d) Nickel (III) sulfate
e) Copper (I) sulfide
f) aqueous hydrogen nitrite
g) ammonium carbonate
h) Hydro-iodic acid
i) Calcium hypochlorite
a) Hg3N2
d) Ni2(SO4)3
g) (NH4)2CO3
b) Ca(NO2)2
e) Cu2S
h) HI(aq)
c) Fe(C2H3O2)3
f) HNO2(aq)
i) Ca(ClO)2
10) Identify the shape of the following molecules and characterize them as “polar” or “non-polar.”
If the molecule is polar, identify the side that would be slightly negative.
a) NH3 b) CCl4 c) PF3 d) CO2 e) HCN f) HCF3 g) BH3 h) HI i) O2 j) H2O k) PH3
a) trigonal pyramidal (3 bonds, l lone pair); polar (asymmetrical, polar bonds)
b) tetrahedral (4 bonds, 0 lone pairs); non-polar
c) trigonal pyramidal (3 bonds, l lone pair); polar (asymmetrical, polar bonds)
d) linear (2 bonds, 0 lone pairs); non-polar (polar bonds, but symmetrical)
e) linear (2 bonds, 0 lone pairs); polar (polar bonds, symmetrical shape, but bonds not the same)
f) tetrahedral (4 bonds, 0 lone pairs); polar (polar bonds, symmetrical shape, but bonds not the same)
g) planar triangular (3 bonds, 0 lone pairs); non-polar (bonds not polar)
h) diatomic; polar (bond is polar)
i) diatomic; non-polar (bond is not polar)
j) bent (2 bonds, 2 lone pairs); polar (polar bonds and asymmetrical shape)
k) trigonal pyramidal (3 bonds, l lone pair); non-polar (bonds not polar)
11) Answer the following:
a) Why is F2 a gas when I2 is a solid? (I2 has a more electrons for London force)
b) Why is NH3 soluble and PH3 not? (NH3 is polar and can Hydrogen bond, PH3 is not polar)
c) Why is the boiling point of NH3 so much higher than that of NF3
(NH3 can Hydrogen bond, NF3 cannot)
d) Why is the boiling point of NH3 so much higher than PH3
(NH3 is polar and can Hydrogen bond, PH3 is not polar)
e) Why is the boiling point of NH3 so much higher than BH3
(NH3 is polar and can Hydrogen bond, BH3 is not polar)
f) Why is the melting point of NaCl so much higher than KBr
(The ions in NaCl are smaller, so the ionic charges are more concentrated and, thus, stronger)
g) Why is CH2F2 polar and CH4 not
(The bonds in CH2F2 are polar, and not all the same)
h) Why is C2H4 planar while C2H6 is not
(Each Carbon in C2H2 is trigonal planar, but is tetrahedral in C2H6)
12) i) Draw structural diagrams for the following molecules.
ii) Identify any co-ordinate covalent bonds present.
iii) Show how hydrogen ions would bond to the molecules in ‘d’ and ‘e’ to form neutral molecules.
iv) How would the ions in ‘d’ and ‘e’ form in the first place?
a) NH4+ b) SO2 c) SO3 d) SO32- e) SO42- f) C2H6O
ii) Co-ordinate bonds are present in a, b, c, d, e
(single bonded oxygens for b and c, non-charged single bonded oxygens for d and e)
iii) Attach H atoms to 2 of the oxygens in each case.
iv) They form when acids lose the Hydrogens attached to the oxygen atoms.
13) Balance the following skeleton chemical equations.
a) 2H2O2 —> 2H2O + O2
b) 2Na3PO4 + 3CaCl2 —> 6NaCl + Ca3(PO4)2
c) 4NH3 + 3O2 —> 2N2 + 6H2O
d) 2C5H6 + 13O2 —> 10CO2 + 6H2O
14) Predict the products of the following reactions and balance the resulting chemical equation.
If there is no reaction, write ‘NR.’
a) P4 + 5O2 —> 2P2O5
b) 3Zn(s) + 2FeCl3(aq) —> 3Fe(s) + 3ZnCl2(aq)
c) 3HCl(aq) + Al(OH)3(aq) —> AlCl3(aq) + 3H2O(l) d) KBr(aq) + (NH4)2SO4(aq) —> NR (all are soluble)
e) CaCO3 —> CaO(s) + CO2(g)
f) 2Ag + Cl2 —> 2AgCl
g) Pb(s) + Mg(NO3)2(aq) —> NR (Pb is less reactive) h) Na2CO3(aq) + CaCl2(aq) —> CaCO3(s) + 2NaCl(aq)
i) 2NaHCO3 —> Na2O(s) + 2CO2(g) + H2O(g)
15) If you had 6.5 g of liquid Nitrogen and allowed it to evaporate and fill a balloon as Nitrogen gas:
a) How many Nitrogen molecules would you have? (6.5g/28.02 g/mol) x 6.02x1023 = 1.4x1023
b) How many Nitrogen atoms would you have? 2.8x1023
c) Why are these two numbers not the same? There are 2 Nitrogen atoms per Nitrogen molecule.
16) For the following reaction: 3Mg(s) + 1N2(g) —> 1Mg3N2(s)
a) If 109.5 g of Mg and 70.00 g of N2 gas are reacted in this manner, determine the limiting reactant.
109.5 g / 24.31 g/mol = 4.504 mol; 4.504 mol / 3 = 1.501 <== smaller number means limiting reactant
70.00 g / 28.02 g/mol = 2.498 mol; 2.498 mol / 1 = 2.498
b) Using the data from part ‘a,’ what mass of Magnesium nitride should be produced?
start with moles of limiting reactant: 4.504 mol / 3 x 1 = 1.501 mol of Magnesium nitride
1.501 mol x ((3 x 24.31 g/mol)+(2x14.01g/mol)) = 1.501 mol x 100.95g/mol = 151.5 g of Magnesium nitride
c) If the reaction in part ‘a’ took place and 148.5 g of Mg3N2(s) is collected, what is the percent
yield?
percent yield = actual yield/theoretical yield x 100% = 148.5g/151.5gX100% = 98.02%
17) Write an ionic equation for the dissolution of Ag3PO4(s) in water.
Ag3PO4(s) —> 3Ag+(aq) + PO43-(aq)
18) A stock solution of HCl(aq) has a molar concentration of 16.0 mol/L. What volume of this stock
solution is needed to prepare 750.0 mL of 0.250 mol/L solution. Describe how you would accurately
prepare this dilute solution.
Make certain, when using this equation, that your units are consistent (epspecially volume, which can be in L or mL)
nc = nd
CcVc=CdVd
Vc=CdVd/Cc = 0.250 M x 0.7500 L / 16.0 M = 0.0117 L = 11.7 mL
19) For the reaction between Lead (II) nitrate and Potassium chloride:
Pb(NO3)2(aq) + 2KCl(aq) —> PbCl2(s) + 2KNO3(aq)
a) If you want to completely react 100.0 mL of 0.50 mol/L lead (II) nitrate solution, what volume of
0.35 mol/L Potassium chloride will you need?
moles of lead (II) nitrate used:
moles of KCl(aq) needed:
volume of KCl(aq) needed:
n = CV = 0.50 M x 0.1000 L = 0.050 mol
nPb(NO3)2 /1 x2 = 0.050 mol x 2 = 0.10 mol
V = n/C = 0.10 mol/0.35 mol/L = 0.29 L = 290 mL
b) If 55.5 mL of 0.25 mol/L Potassium chloride completely reacts, what mass of Lead (II) chloride
should be produced?
moles of KCl(aq) used:
moles of PbCl(s) made:
mass of PbCl(s) made:
n = CV = 0.25 M x 0.0555 L = 0.01388 mol
nKCl /2 x1 = 0.006938 mol
m = n x mN = 0.006938 mol x (207.20 g/mol + (2 x 35.45 g/mol))
= 0.006938 mol x 278.1 g/mol
= 1.9 g
20) A solution of Lithium hydroxide is titrated using a standard solution of 0.150 mol/L sulfuric acid.
2LiOH(aq) + H2SO4(aq) —> 2H2O(l) + Li2SO4(aq)
a) If 35.0 mL of the acid are needed to neutralize 15.0 mL of the base, what is the concentration of
the base?
nB = nA /1 x2 = CAVA /1 x2 = 0.150 M x 0.0350 L /1 x2 = 0.0105 mol
concentration of base: CB=nB/VB = 0.0105 mol / 0.0150 L = 0.700 M (remember that M is mol/L)
b) Write an ionic and net ionic equation for this reaction.
2LiOH(aq) + H2SO4(aq) —> 2H2O(l) + Li2SO4(aq)
i.e.
2Li+(aq) + 2OH-(aq) + 2H+(aq) + SO42-(aq) —> 2H2O(l) + 2Li+(aq) + SO42-(aq)
net i.e. OH-(aq) + H+(aq) —> H2O(l)
shows actual form of each substance
shows only substances that change form
21) Identify a set of test solutions and a correct sequence for adding them to an unknown solution to
test for the presence of one or more of the following metal ions: Mercury (2+), Silver, and Barium
Use table 1 at the end of the review.
1) Acetate ion (as aqueous Sodium acetate) will precipitate Ag+(aq) if it is present, since AgC2H3O2 has a low solubility.
Only acetate will precipitate the silver ion without precipitating either of the other two ions.
2) Once all the silver ions have been removed, if they are present, we can test for one of the other two ions.
a) We could add hydroxide ion to test for Mercury 2+ (it won’t precipitate Barium) or
b) add sulfate ion to precipitate Barium as it won’t precipitate Mercury 2+.
3) Once the second ion has been removed, if present, use the final test for the third ion.
a) if you used step 2a, add sulfate, carbonate, or phosphate to test for Barium ion.
b) if you used step 2b, add hydroxide, carbonate, or phosphate to test for Mercury 2+ ion.
22) What is the concentration of sulfate ions in a 2 mol/L solution of Aluminum sulfate?
Since the formula for Aluminum sulfate is Al2(SO4)3, the concentration will be 2 mol/L x 3 = 6 mol/L
23) Identify the type of solution (unsaturated, saturated,
supersaturated) for the following in the graph to the right:
a) 0.7 mol/L @ 70 oC
saturated
o
supersaturated
b) 0.4 mol/L @ 34 C
c) 1.0 mol/L @ 90 oC
unsaturated
24) Describe what will take place as a 0.75 mol/L solution @
85 oC is cooled to 20 oC while being agitated.
Draw a point at 0.75M and 85oC.
Extend a horizontal line to the left until it meets the curve.
Extend a line along the curve to the left until it reaches 20 oC.
The concentration will be 0.2 mol/L at that temperature.
At that point, 0.55 moles of solute will have precipitated out per litre of solution (0.75 - 0.20).
25) Calculate the pH of the following solutions:
a) .025 mol/L of HCl(aq), b) 0.025 mol/L of H2SO4(aq),
c) 3 mol/L of acetic acid (strength is 13%)
a) -log(1 x 1 x 0.025) = 1.6
b) -log(1 x 2 x 0.025) = 1.3
c) -log(0.13 x 1 x 3) = 0.41
26) Give the Hydronium ion concentration in the following solutions:
a) pH = 3, b) pH=12.5, c) pOH = 5
a) 10-3 M
b) 10-12.5 = 3.16 x 10-13 M
c) 10-(14-5) = 10-9 M
27) a) Why is the pH of a 0.1 mol/L solution of HNO3 equal to 1.0 when the pH of a 0.1 mol/L
solution of HCN is 5.0?
HNO3 is a strong acid, while HCN is a very weak acid (most of the HCN molecules do not ionize in solution).
b) Express the difference between the two acids quantitatively.
If an acid is strong, than a 0.1 M solution should have a pH of 1.0 (100% strength)
0.1 M HCN has a pH of 5.0, which means that its H+ concentration is 10-5 M
strength = 10-5M / 0.1M x 100% = 0.01%
28) a) Give Arrhenius and Brønsted-Lowry definitions for acids and bases.
Acids donate Hydrogen ions, bases accept them.
Neutral substances do neither.
Amphoteric substances can do either.
b) What improvements did the Brønsted-Lowry theory make compared to the Arrhenius theory?
The Arrhenius theory could not explain amphoteric substances and did not explain acid/base properties for situations that
did not involve aqueous solutions.
29) Identify the Brønsted-Lowry acid, base, conjugate acid, and conjugate base for the following
reaction: NaHSO4(aq) + NaHCO3(aq) —> Na2SO42-(aq) + H2CO3(aq)
acid
base
conj. base
conj. acid
30) Write chemical equations (with HCl and with NaOH) to show that NaHCO3 is amphoteric (can act
as an acid or base).
NaHCO3(aq) + HCl(aq) —> H2CO3(aq) + Na+(aq) + Cl-(aq)
(H2CO3 is unstable and decomposes to CO2 and H2O which causes the familiar fizzing)
NaHCO3(aq) + NaOH(aq) —> 2Na+(aq) + CO32-(aq) + H2O(l)
31) a) If a sample of gas at constant temperature experiences a tripling of the ambient pressure, what
will happen to the volume of the gas sample?
P1V1=P2V2
V2 = P1V1/P2
Since P1/P2 = 1/3, the new volume, V2, will be 1/3 of V1
V1/T1 = V2/T2
V2 = V1 x T2 / T1 Since T2/T1 is 4, the new volume, V2, will be 4 times larger than V1.
b) If the absolute temperature of a gas is quadrupled while the pressure remains constant, what will
happen to the volume of the gas?
32) A balloon at Standard Pressure has a volume of 2.50 L. The balloon is placed in a pressure
chamber at constant temperature and the volume is reduced to 0.850 L. What is the pressure in the
chamber?
Pressure at STP is 101.3 kPa. This is P1. V1 is 2.50 L
P2 = P1V1/V2 = 101.3 kPa x 2.50 L / 0.850 L = 298 kPa
33) A weather balloon is filled with Helium gas at a temperature of 35.5 oC and a pressure of 102.5
kPa. The volume of the balloon is 50.0 L. The balloon is constructed so that it can expand and
contract easily. When the balloon is released and reaches the Stratosphere, the pressure has fallen to
30.1 kPa and the temperature has fallen to -40.5 oC. What is the volume of the balloon at this altitude?
P1V1/T1 = P2V2/T2
V2 = P1V1T2/(T1V2) = 102.5 kPa x 50.0 L x (35.5 + 273.15) / (30.1 kPa x (-40.5 + 273.15))
= 102.5 x 50.0 x 308.65 / (30.1 x 232.65)
= 226 L
34) What is the volume of 0.250 moles of N2 gas at 155 oC and 175 kPa?
PV=nRT
V = nRT/P = 0.250 mol x 8.314 x (155 + 273.15) / 175 = 5.09 L
35) 3 moles of Argon gas, 4 moles of Neon gas, and 5 moles of Helium gas are placed in a rigid tank at
SATP. Calculate the volume of the tank and the partial pressure of each of the gases in the mixture.
P @ SATP = 100.0 kPa
T @ SATP = 298.15 K
Any ideal gas will behave just like any other ideal gas. Thus, we can say that we have 12 moles of gas in total.
V = nRT/P = 12 x 8.314 x 298.15 / 100 = 297 L or about 300 L, (we only have 1 significant digit for our mole amounts).
To calculate partial pressure, use one of two techniques:
a) Px = nxRT/V
b) Px = Ptotal(nx/ntotal)
PAr = 100(3/12) = 25 kPa
PNe = 100(4/12) = 33 kPa
PHe = 100 - 25 - 33 = 42 kPa
36) At STP, a 7.50 g sample of an unknown gas has a volume of 3.82 L.
a) What is the density of the gas? Dgas = 7.50 g / 3.82 L = 1.96 g/L
Note: there was a mistake in the original question: air density is 1.29 g/L, not 1.29 g/mL (water is only 1g/mL!)
b) If air at STP has a density of 1.29 g/L, would the gas float or sink in air? It will sink (Dgas>Dair).
c) How many moles of gas is present in the sample? n=PV/RT = 101.3 x 3.82 / (8.314 x273.15) = 0.170 mol
d) What is the molar mass of this gas? mN = m/n = 7.50 g / 0.170 mol = 44.0 g/mol
e) Which of the following is the identity of this gas: HCl, O2, CO2, HCN?
Since the molar mass of CO2 is 44.01 g/mol, the mystery gas must be CO2
37) Nitrogen tri-iodide will explode at the slightest contact to produce Nitrogen gas and Iodine gas in
the following reaction: 2NI3(s) —> N2(g) + 3I2(g)
If 15.5 g of NI3 explodes at a temperature of 25.5 oC and a pressure of 100.3 kPa.
a) What volume of gas will be produced? We will ignore part b for this question, I don’t like much anymore
nNI3 = 15.5 g / (14.01 g/mol + (3 x 126.90 g/mol)) = 15.5 g / 394.71 g/mol = 0.0393 mol
nN2 = nNI3 /2 x1 = 0.0393 /2 = 0.0196 mol
V= nRT/P = 0.0196 x 8.314 x (25.5+273.15)/100.3 = 0.486L
V=nRT/P = 0.0590 x 8.314 x (25.5+273.15)/100.3 = 1.46L
nI2 = nNI3 /2 x3 = 0.0393 /2 x3 = 0.0590 mol
The total gas volume is 1.46 L + 0.486 L = 1.94 L
38) Answer the following:
a) For Ca2+ and Cl-, compare the radii of these ions and explain the difference.
Both have the same number of electron shells, but Ca2+ has fewer electrons per proton, so it is smaller (more attraction to
nucleus).
b) Why is the second ionization energy of K greater than that of Ca while the first ionization energy
of Calcium is higher than that of Potassium?
Since Potassium has only one electron in its outer shell, the second electron must be removed from the next shell in, which
takes a lot more energy. Calcium has two electrons to remove from the outer shell. Because Calcium has more protons, it
is harder to remove electrons from its outer shell than it is for Potassium.
c) Why is methane gas not ideal at very low temperatures and high pressures?
London forces are able to attract gas molecules together at low temperatures and high pressure.
d) Why do water droplets form on the outside of containers of ice water?
When water vapour molecules collide with the container, they lose more kinetic energy than when they collide with warmer
surfaces (the molecules of a warm container have more kinetic energy themselves, so they take less energy from the
colliding gas molecule). Thus, the water vapour molecule is more likely to lose too much energy to stay in the vapour state.
e) Why do structures made of calcium carbonate deteriorate quickly in areas with acid rain?
Carbonate is a base, when it accepts 2 Hydrogen atoms and becomes carbonic acid, this decomposes into Carbon dioxide
and water. The CO2 escapes into the atmosphere. With less acidic rain, this process is much slower.
39) For the following reaction: 4Fe(s) + 3O2(g) —> 2Fe2O3(s), 75.0 g of Iron is reacted with 11.5 L of
Oxygen at 2.66atm and 25 oC. Calculate the theoretical yield of iron III oxide.
75.0 g Fe / 55.85 g/mol = 1.34 mol 1.34 mol / 4 = 0.336
11.5 L x 2.66 atm / (0.82 x 298.15) = 0.125 mol 0.125 mol / 3 = 0.0417 <=== limiting reactant. (R=0.82 when P in atm)
0.125 moles of O2 /3 x2 = 0.0835 moles of Iron (III) oxide should be made.
the mass of Iron (III) oxide is 0.0835 moles x ((2 x 55.85g/mol) + (3 x 16.00 g/mol)) = 13.3 g
40) Write balanced equations, and where appropriate ionic, and net ionic equations, for the following:
a) Iron nitrate is spilled onto a sheet of Aluminum Al(s) + Fe(NO3)3(aq) —> Al(NO3)3(aq) + Fe(s)
b) C6H12O6 is completely combusted in air
c) HCl is added to a solution of carbonate
Al(s) + Fe3+(aq) + 3NO3-(aq) —> Al3+(aq) + NO3-(aq) + Fe(s)
Al(s) + Fe3+(aq) —> Al3+(aq) + Fe(s)
C6H12O6(s) + 6O2(g) —> 6CO2(g) + 6H2O(g)
HCl(aq) + Na2CO3(aq) —> H2CO3(aq) + NaCl(aq)
H+(aq) + Cl-(aq) + 2Na+ + CO32-(aq) —> H2CO3(aq) + Na+(aq) + Cl-(aq)
H+(aq) + CO32-(aq) —> H2CO3(aq)
d) Magnesium pellets are placed in 1.0M aqueous hydrogen chloride
Mg(s) + 2HCl(aq) —> MgCl2(aq) + H2(g)
Mg(s) + 2H+(aq) + 2Cl-(aq) —> Mg2+(aq) + 2Cl-(aq) + H2(g)
Mg(s) + 2H+(aq) —> Mg2+(aq) + H2(g)
+ 8O2(g) —> 8SO2(g)
e) Sulfur is burned in air S8(s)
f) solutions of silver nitrate and sodium acetate are combined.
AgNO3(aq) + NaC2H3O2(aq) —> AgC2H3O2(s) + NaNO3(aq)
Ag+(aq) + NO3-(aq) + Na+(aq) + C2H3O2-(aq) —> AgC2H3O2(s) + Na+(aq) + NO3-(aq)
Ag+(aq) + C2H3O2-(aq) —> AgC2H3O2(s)
g) Zinc carbonate is heated strongly ZnCO3 —> ZnO(s) + CO2(g)
h) Calcium hydrogen carbonate is heated strongly Ca(HCO3)2(s) —> CaO(s) + CO2(g) + H2O(g)
i) Barium oxide is mixed with water BaO(s) + H2O(aq) —> Ba(OH)2(s)
j) Sulfur trioxide is mixed with water SO3(g) + H2O(l) —> H2SO4(l)
Table 1: Solubility @ SATP
Cations
Anions
High Solubility
Cl- Br- I-
S2-
OH-
SO42-
CO32- PO43-
CH3COO-
NO3-
most
group 1
group 2
group 1
most
group 1
NH4+
most
all
Ag+ Pb2+
Ca2+ Ba2+
Sr2+ Ra2+
most
Ag+
none
Ag+ Pb2+
Hg2+ Cu+
Tl+
Low Solubility
NH4+
NH4+ Sr2+
Ba2+ Tl+
most
most
Table 3: Flame Test Colours
Table 2: Solution Colours
Table 3: Reactivity Series for Metals
Ion
Colour
Ion
Colour
groups 1, 2, 17
colourless
H+
colourless
Lithium
Cr2+
blue
Li+
bright red
Potassium
Cr3+
green
Na+
yellow
Barium
violet
Calcium
yellow-red
Sodium
+
+
Cu
blue
K
Cu2+
green
Ca2+
2+
2+
Most Reactive
pale green
Sr
bright red
Magnesium
Fe3+
yellow-brown
Ba2+
yellow-green
Aluminum
Ni2+
green
Cu2+
blue (halides)
green (others)
Zinc
Pb2+
light blue-grey
Zn2+
whitish green
Fe
2+
pink
Co
Gas Law Data
STP
0oC
o
Iron
Tin
Lead
Hydrogen
101.3 kPa
Copper
SATP
25 C
100 kPa
Silver
Ideal Gas
Constant
8.31
0.082
kPaAL/molAK
atmAL/molAK
Gold
Least Reactive