James Richard Fromm
1. Name, using correct modern chemical nomenclature, each of the compounds whose formula is given. Give the correct molecular formula for each of the named compounds; if no molecular formula exists, give the correct empirical formula: a, Cr2O3(s); b, LiCO3(s); c, Co(NH3)63+(aq);
d, Ca(BrO4)2(s); e, K[Al(OH)4](s); f, sodium dihydrogen phosphate; g, hypochlorous acid; h, vanadium (V) chloride; i, potassium dicyanoargentate (I); j, silver sulfite.
2. a. Write both the form and the value (in molar units) of the base ionization
constant of dichloroacetate ion:
b. Write both the Lewis structure of the sulfur atom and the Lewis structure of the thiosulfate ion:
c. Write the form of the stability constant of the Cd(CN)66- ion, the tetracyanocadmate(II) ion. In a solution whose concentration of cyanide ion were 0.01 molar, you would expect most of the soluble cadmium (II) to be in the form of (Cd2+ ions, Cd(CN)64- ions, metallic cadmium).
d. Write both the form and the value (in molar units) of the acid ionization constant of arsenic acid, H3AsO4:
e. The complete electronic structure of an iron atom would be written as 1s2 _____________________________. The lowest oxidation state normally observed for iron is Fe(II), and the complete electronic structure of this ion is 1s2_________________________.
3. a. A chemist prepared solid silver(I) sulfide, Ag2S, by reacting together
silver metal and sulfur. He took 46.21 g of silver and 18.43 g of sulfur. If the reaction
went to completion he should have produced ____________ g of silver sulfide and had
____________ g of ____________ (silver, sulfur) remaining.
The actual amount of silver sulfide he produced was 51.63 g, so his percentage yield was ____________%. This actual amount of silver sulfide could be converted to a maximum of ____________ g of silver(I) sulfate, Ag2SO4.
b. The solubility of silver sulfide in water is relatively low. Calculate the molar solubility of Ag2S in distilled water. The molar solubility is, in mol/L:
c. As the temperature increases, the molar solubility of Ag2S should ____________ (increase, decrease).
4. a. Upon heating of a sample of dry mercury(II) oxide, oxygen gas was produced by
Antoine Lavoisier. If in one of his experiments he produced 136.24 mL of gas at a
temperature of 22.4oC by this procedure over water when the atmospheric
pressure was 756.43 torr, calculate the amount of gas that he produced. The amount of
oxygen gas was, in mol of O2(g):
b. Calculate the mass of HgO required to produce the oxygen observed in this experiment. The mass of HgO required is, in g:
c. The oxygen produced by Lavoisier was studied by one of his colleagues. In one experiment, 0.452 moles of oxygen gas were reacted with 8.93 grams of carbon producing heat and carbon dioxide. Calculate the amount of heat produced by this reaction at 25oC and one atmosphere pressure. The amount of heat produced was __________________ kJ per mole of _______________ and thus would be a total of _______________ kJ.
5. a. When the total pressure on a sample of gas which is a mixture of H2(g),
I2(g), and HI (g) is increased, the relative partial pressure of HI(g) will
(increase, decrease, change very little).
b. When the temperature of this sample of gas (part a) is increased but the total pressure is not changed, the relative partial pressure of HI(g) will (increase, decrease, change very little).
c. When the total pressure on a sample of gas which is a mixture of H2(g), O2(g), and H2O(g) is increased, the relative partial pressure of H2O(g) will (increase, decrease, change very little).
d. When the temperature of this sample of gas (part c) is increased but the total pressure is not changed, the relative partial pressure of H2O(g) will (increase, decrease, change very little).
e. When the temperature of the reaction in which gaseous HF(g) dissolves in water and ionizes to produce H+(aq) and F-(aq) is decreased, we would expect the concentration of fluoride ion to (increase, decrease, remain much the same).
6. Acme Gas Ltd. is designing a plant to produce the hydrogen required for ammonia synthesis from natural gas. The overall reaction is CH4(g) + O2(g) CO2(g) + 2H2(g), although in the actual process several steps are required to accomplish this process. For the overall reaction, the equilibrium constant calculated at 25oC from standard thermodynamic data is 1.57 x 10+60 bar. If the reaction were the overall equilibrium reaction, what are the equilibrium partial pressures of unreacted methane and oxygen if the total pressure is 60 bar and if the reaction were originally fed with a 1:1 stoichiometric mixture of methane and oxygen (in bar)?
7. a. A white precipitate of aqueous mercury(II) hydroxide was formed when an aqueous
solution of Hg(NO3)2 had added to it sufficient KOH that the
equilibrium concentration of hydroxide ion was 0.0145 molar. Compute the concentration of
Hg2+(aq) remaining in the solution at equilibrium. The concentration of Hg2+(aq)
was, in mol/L:
b. Compute the molar solubility of Hg(OH)2 in distilled water. The molar solubility of Hg(OH)2 in distilled water at 25oC is ____________ mol/L. This value (is, is not) the same as the molar concentration of Hg2+(aq) and (is, is not) the same as the molar concentration of hydroxide ion.
c. Sufficient hydrochloric acid is added to change the solution pH of the solution in part a above to 7.23 as measured by a pH meter. Compute the concentration of Hg2+(aq) in solution now, and then explain the difference (if any) between this result and the result you gave in part a on the basis of LeChatelier's principle. The value is ___________________ mol/L and my explanation is:
8. A chemist wished to measure the formic acid content of an ant and used the following
procedure. To one crushed ant he added 35.46 mL of 0.0116 molar sodium hydroxide and
titrated the resulting solution with 39.45 mL of 0.0100 molar standard HCl.
a. Assuming ants contain no other acids, the formic acid content of the ant, is, in mol/ant:.
b. Sketch the titration curve for this titration, indicating any buffer regions present.
c. Calculate the pH of an aqueous buffer made from 0.0169 moles of formic acid and 0.00692 moles of sodium hydroxide. The pH of such a buffer is:
9. A certain FM station broadcasts on 92.5 MHz. For this station the wavelength is ____________ nm and the energy of one photon is __________ J. The station uses 50,000 W (1 W = 1 J/s) so the photonic output of this station is ____________ photons/s or ____________ mol photons/s.
The entire output of this station is channeled by appropriate radio receivers into one 70 kg student. Assuming that the student consists entirely of hydrogen atoms and that each photon interacts with one hydrogen atom, it will take ____________ s for every atom of the student to reach a (higher, lower) energy state. Once this state is reached, the electrons in all these hydrogen atoms will be (closer to, farther from) their nuclei. These electrons can then resume their original states by (emitting, absorbing) photons of (the same, greater) energy than the photons emitted by the FM station.
If the photons emitted by this station did not correspond in energy to the difference
between two orbits for an electron in hydrogen, as we assumed in the above paragraph, the
photons (would, would not) interact with the hydrogen atom and would then (remain
unaffected, be absorbed, be emitted, be destroyed) when they encounter a hydrogen atom.
Show your calculations here:
10. a. A neutral atom has the atomic structure 1s22s22p63s23p64s23d5.
The atom is an atom of the element:
b. Give the complete electronic structure of an atom of rubidium (Rb):
c. The principal quantum number of an electron orbital is 3 and its momentum quantum number is 0. Give all of the permissible values of the two remaining quantum numbers: m = _______________, s = _______________.
d. The bonding in carbon monoxide should be (more, less) polar than the bonding in nitrogen monoxide.
e. A covalent bond formation between two atoms is the result of the (transfer, sharing, interference, destruction) of electrons between them.
f. Draw the shape of the PCl5 molecular according to the V.S.E.P.R. approach:
g. Draw the shape of the PCl3 molecule according to the V.S.E.P.R. approach:
h. Draw the Lewis structure of nitrogen monoxide:
i. The shape of the Fe(CN)63-(aq) ion is called:
j. The shape of the CH4(g) molecule is called:
1. a. chromium (III) oxide or dichromium trioxide; b. lithium carbonate; c. the tetraammine- cobalt (III) ion; d. calcium bromate or calcium tetraoxobromate(VII); e. Potassium tetra- hydroxoaluminate(III); f. NaH2PO4; g. HClO; h. VCl5; i. K[Ag(CN)2]; j. Ag2SO3.
2. a. [OH-][Cl2HCCOO-]/[Cl2HCCOH], 2 x 10-13; b. :S:: and ::S=S (= O::)3,
c. [Cd(CN)64-]/[Cd2+][CN-]4, Cd(CN)64- ions; d. [H+][H2AsO4-]/[H3AsO4], 5.6 x 10-3;
e. 2s22p63s23p64s23d6, 2s22p63s23p64s03d6
3. a. 53.08 g of silver sulfide and had 11.56 g of sulfur remaining.
2Ag(s) + S(s) Ag2S(s)
46.21 g/107.868 g/mol = 0.4284 mol Ag
18.43 g/32.06 g/mol = 0.5749 mol S
At 2:1 Ag is limiting, so 0.2142 mol Ag2S, which is 53.08 g Ag2S.
The amount of S left = 0.5749 - 0.2142 = 0.3607 mol
0.3607 mol x 32.06 g/mol S = 11.56 g S left over
51.63 g/53.08 g x 100 = 97.3% yield
51.63 g/247.796 g/mol = 0.2084 mol
Ag2S which can react to form not more than 0.2084 mol Ag2SO4
0.2084 mol Ag2SO4 x 311.796 g/mol = 64.96 g
b. 1.4 x 10-17
Ag2S(s) 2Ag+(aq) + S2-(aq)
Ksp = [Ag+]2[S2-] = 1 x 10-50
[Ag+] = 2[S2-] from stoichiometry 1 x 10-50 = (2[S2-])2[S2-] = 4[S2-]3; [S2-]3 = 0.25 x 10-50, [S2-] = 1.36 x 10-17
c. increase. For most salts this is true. You don't have the data you need, DHof(Ag2S), to calculate the enthalpy change for this reaction.
4. a. 5.44 x 10-3
pV = nRT, V = 136.24 mL; T = 295.55 K; p = 756.43 - 20.3 (interpolated) = 736.13 torr, which is 736.13/760 = 0.969 atm
n = 0.969 atm x 0.13624 L/0.08206 L atm/K mol x 295.55 K = 5.441 x 10-3 mol
b. 2.36 g 2HgO 2Hg + O2; 2 mol HgO per mol O2
molar mass = 200.59 + 16.00 = 216.59 g/mol HgO
5.44 x 10-3 mol O2 x 2 x 216.59 g/mol = 2.356 g HgO
c. -394 kJ per mole of O or CO2 and thus would be a total of -178 kJ.
C(s) + O2(g) CO2(g)
0.743 mol C + 0.452 mol O2 0.452 mol CO2 (O2 limiting)
8.93/12.011 = mol C = 0.743 mol C
DH0 = -394 - (0+0) = -394 kJ/mole reaction or mol C or mol O2
0.452 x -394 = -178.1 kJ of heat
5. a. change very little. H2(g) + I2(g) 2HI(g). Same number of moles of gas on
both sides, so pressure has no significant effect.
b. increase. H2(g) + I2(g) 2HI(g). DH0 = 2DH0f(HI) - 2(0) = 2 x 26 = +52 kJ; reaction is endothermic: H2 + I2 + heat --> 2HI. The relative partial pressure of HI will therefore INCREASE.
c. increase. 2H2(g) + O2(g) 2H2O(g); moles of gas are less on the product side, so an increase in pressure produces relatively more products.
d. decrease. 2H2(g) + O2(g) 2H2O(g), DH0 = 2DH0f(H2O) - 2(0) - (0) = 2(-242) = -484 kJ; exothermic: 2H2 + O2 --> H2O + heat. Addition of heat drives it backwards so relative partial pressure of water vapor decreases.
e. increase. HF(g) H+(aq) + F-(aq). DH0 = 0.0 + (-333) - (-271) = -52 kJ, exothermic; HF(g) H+(aq) + F-(aq) + heat. Addition of heat drives it backwards, so decrease in temperature would drive it forward and then the concentration of fluoride ion would increase.
6. 1.4 x 10-28 bar and the partial pressure of oxygen is 1.4 x 10-28
bar at equilibrium. Alternatively use the quadratic equation. Essentially all of it
Kp = p(CO2)p2(H2)/p(CH4)p(O2) = 1.57 x 10+60
The stoichiometry of the feedstock is 1:1 CH4:O2 so the initial partial pressures are equal. The reaction consumes these in 1:1 ratio so the partial pressures are always equal. Then Kp = p(CO2)p2(H2)/p2(CH4). There is originally no CO2 and no H2 present. They are formed only by the reaction in the ratio 2H2:1CO2. The ratio of partial pressures is then always p(H2) = 2p(CO2).
Kp = p(CO2)[2p(CO2)]2/p2(CH4) = 4p3(CO2)/p2(CH4)
p = 60 bar = p(CO2) + p(H2) + p(CH4) + p(O2)
60 bar = p(CO2) + p(H2) + 2p(CH4)
60 bar = p(CO2) + 2p(CO2) + 2p(CH4) = 3p(CO2) + 2p(CH4)
60 bar is about 3p(CO2), so p(CO2) is about 20 bar, due to high value of Kp
p2(CH4) = 4(8000)/1.57 x 10+60 = 32000/1.57 x 10+60 = 2.04 x 10-56 bar2
p(CH4) = 1.43 x 10-28 bar
7. a. 1.4 x 10-22 mol/L
Ksp = [Hg2+][OH-]2 = 3 x 10-26
[Hg2+] = 3 x 10-26/(0.0145)2 = 1.43 x 10-22
At pH 7.23, pOH = 6.77 and [OH-] = 1.70 x 10-7
[Hg2+] = 3 x 10-26/(1.70 x 10-7)2 = 1.04 x 10-12
b. 2 x 10-9 mol/L. This value is the same as [Hg2+(aq)], and is not the same as [OH-].
Ksp = [Hg2+][OH-] = 3 x 10-26
There is no other ion source in distilled water, so the stoichiometry is 2OH-:1Hg2+ and [OH-] = 2[Hg2+]. Then 3 x 10-26 = 4[Hg2+]3, [Hg2+] = 1.96 x 10-9 mol/L.
c. 1.0 x 10-12 mol/L. Addition of acid removes hydroxide to form water, so the concentration of hydroxide decreases. The hydroxide would increase as more Hg(OH)2 dissolved, so the equilibrium is following Le Chatelier's principle.
8. a. 1.68 x 10-5 mol formic acid/ant. 35.46 x 0.0116 = 0.4113 mmol NaOH;
39.45 x 0.0100 = 0.3945 mmol HCl; difference = 0.0168 mmol formic acid = 1.68 x 10-5
mol formic acid
b. The titration curve is plotted with pH as the y (vertical) axis and volume of HCl added as the x (horizontal) axis, from zero onwards. The sketch has a long plateau running at about pH 12 until nearly the equivalence point, then drops to a very small plateau near pH 4 and finally to a long plateau about pH 2 due to the excess HCl. The very small plateau is due to the formic acid/formate buffer system (the ant).
c. The pH of such a buffer is 3.86.
Ka = [H+][formate-]/[Hformate] = 2 x 10-4
HCOOH + OH- HCOO- + H2O
0.01690 mol HCOOH - 0.00692 mol NaOH = 0.00998 mol HCOOH left; reaction produces 0.00692 mol HCOO-
2 x 10-4 x 0.00692/0.00998 = [H+] = 1.39 x 10-4. pH = -log[H+] = 3.858
9. 3.2 x 10+9; 6.1 x 10-26 J, 8.16 x 10+29 photon/s or
1.34 x 10+6 mol photon/s; 0.05 s for every atom of the student to reach a
higher energy state, farther from their nuclei, emitting photons of the same energy; would
not interact, remain unaffected
lf = c = 3 x 10+8 m s-1; f = 92.5 x 10+6 s-1; l = 3 x 10+8 m s-1/92.5 x 10+6 s-1 = 3.243 m = 3.2 x 10+9 nm
E = hf; h = 6.626 x 10-34 J s; E = 6.6 x 10-34 x 92.5 x 10+6 = 6.13 x 10-26 J. 50,000 J s-1/6.13 x 10-26 J = 8.16 x 10+29 photons/s
8.16 x 10+29/6.023 x 10+23 = 1.354 x 10+6 mol photon/s
One 70 kg student = 70,000 mol of H at 1.0 g/mol H. The station emits 1.34 x 10+6 mol photon/s, so the time will be 70,000/1.34 x 10+6 = 0.052 s.
10. a. Mn (manganese); b. 1s22s22p63s23p64s23d104p65s1; c. m = 0, s = +1/2, -1/2 (if l = 0, m = -0 to +0 = 0); d. more. Electronegativity difference is 1.0 in CO and 0.5 in NO so CO should be more polar; e. sharing; f. 10 electrons around P, 5 pairs, basic shape is trigonal bipyramid, no distortion; g. 6 electrons around P, 3 pairs, basic shape is trigonal plane, no distortion; h. 6 + 5 electrons = 11 electrons; i. octahedral (hexacoordinate is usually octahedral); j. tetrahedral.