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Chapter 1 · Measurement, significant figures and uncertainty
The figure shows five pieces of glassware with the tolerance stated by the manufacturer.
c) An aliquot is to be taken for a dilution: the volume taken will go directly into the calculation of the final concentration. Which of the five instruments should be chosen?
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Answer
c)The 25.00 mL volumetric pipette
c) The volumetric pipette. It is the only instrument on the list designed to DELIVER a single, fixed volume with high precision, which is exactly what an aliquot requires. The volumetric flask, on the other hand, is used to CONTAIN a volume, not to deliver one: the diluted solution is prepared in it, nothing is withdrawn from it.
Chapter 4 · Hybridization and the quantum model of bonding
Pauling resolves the contradiction in two steps: first an electron is promoted, which costs energy, then the valence orbitals are combined mathematically into equivalent orbitals. The energy balance is largely favourable.
c) Explain what sp3 hybridization is and how many hybrid orbitals it produces.
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Answer
c)One 2s and three 2p give FOUR identical hybrids, arranged in a tetrahedron
c) sp3 hybridization is the mathematical combination of the 2s orbital and the THREE 2p orbitals. It produces FOUR identical hybrid orbitals, of equal energy, intermediate between those of s and p, and pointing toward the four corners of a tetrahedron. Each holds one unpaired electron and can form a σ bond. The name sp3 gives the recipe: one part s, three parts p.
d) Three: pentane, a straight chain of five carbons; 2-methylbutane, a chain of four with a methyl group on the second carbon; 2,2-dimethylpropane, a chain of three with two methyl groups on the central carbon, which is then quaternary. There is no fourth possibility: a methyl group on the third carbon of a four-carbon chain gives 2-methylbutane again, seen from the other end, which is the most frequent counting error. Their boiling points, 36, 28 and 10 °C, decrease with branching, the molecule becoming more compact and offering less contact surface.
Chapter 6 · Stoichiometry, limiting reactant and percent yield
Ammonia is synthesized according to N2+3H2→2NH3. 28.0 g of dinitrogen N2 are reacted with 5.00 g of dihydrogen H2. Molar masses (g/mol): N2 28.02; H2 2.02; NH3 17.04.
The diagram below shows the synthesis.
b) Determine the limiting reactant by comparing the ratios coefficientn.
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Answer
b)0.825<1.00: dihydrogen is limiting
b) For N2: 11.00=1.00. For H2: 32.48=0.825. The smallest ratio is that of dihydrogen, so H2 is the limiting reactant: it runs out before the dinitrogen. Once the limiting reactant is used up, adding more excess reactant produces nothing more: the amount of product levels off on a plateau.
Chapter 7 · Gases: Dalton, kinetic theory and real gases
In a mixture, each gas behaves as if it were alone: it exerts the pressure it would have if it occupied the whole volume. The total pressure is simply the sum of these contributions, weighted by the mole fractions.
The diagram below shows the mixture of a) and the setup of c).
c) Hydrogen gas is collected by displacement of water: 250 mL at 22 °C, under an atmospheric pressure of 101.3 kPa. The vapour pressure of water at 22 °C is 2.6 kPa. Calculate the mass of H2 collected.
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Answer
c)P(H2)=98.7 kPa, n=1.006×10−2 mol, that is 20.3 mg
c) The gas collected over water is saturated with vapour: the pressure measured is that of the mixture. So P(H2)=101.3−2.6=98.7 kPa. With T=22+273=295 K and V=0.250 L: n=RTPV=8.314×29598.7×0.250=1.006×10−2 mol. The mass is m=1.006×10−2×2.016=2.03×10−2 g, or about 20 mg.
Chapter 8 · Redox, oxidation numbers and nomenclature
A binary ionic compound combines a metal and a nonmetal. Its name follows the order of the formula: the cation first, then the anion with the ending -ide (oxygen gives oxide).
a) NaCl sodium chloride, CaBr2 calcium bromide, Al2O3 aluminum oxide, K2S potassium sulfide. No numerical prefix appears: the subscripts follow from the charges, so they do not have to be said. Writing calcium dibromide is an error, not a harmless redundancy.
A metal reaches the configuration of the noble gas that PRECEDES it by losing its valence electrons. Ionization energies of sodium: first 496 kJ/mol, second 4562 kJ/mol. The extract of the periodic table below places these five metals.
c) Iron gives Fe2+ and Fe3+. Write the configuration of Fe3+ starting from that of iron, [Ar]3d64s2, and say what makes it particularly stable.
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Answer
c)[Ar]3d5: the d subshell exactly half filled
c) Iron is [Ar]3d64s2. The two 4s electrons are removed first, because they are the outermost despite the filling order, which gives Fe2+ as [Ar]3d6. One more 3d electron gives Fe3+ as [Ar]3d5. Its particular stability comes from this d subshell being exactly HALF filled, five orbitals each holding one unpaired electron of the same spin: it is the same reason that explains the anomaly of chromium in the table of configurations, and it is why iron(III) salts are so common.
In solution, an amount of substance reads as a product C×V. All the stoichiometry of the previous chapter then applies unchanged, provided you never confuse the volume of solution with the volume of solute.
Molar mass of AgCl: 143.32 g/mol. The diagram below shows the titration of b) and the precipitation of c).
a) How would you prepare 500 mL of hydrochloric acid at 0.150 mol/L from a commercial solution at 12.0 mol/L? Describe the procedure.
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Answer
a)Withdraw 6.25 mL and make up to 500 mL in a flask that already contains water
a) Dilution conserves the amount of substance: C1V1=C2V2, so V1=C1C2V2=12.00.150×0.500=6.25×10−3 L, or 6.25 mL. Procedure: draw 6.25 mL of the commercial solution with a pipette, transfer it to a 500 mL volumetric flask that ALREADY contains some distilled water, then fill up to the calibration mark and mix. Never put the acid into an empty flask that you then fill with water.
A system exchanges energy with its surroundings in only two ways: as heat and as work. The first law states that the sum of the two is a state function, the internal energy.
Reminder of units: 1 kPa·L = 1 J.
The diagram below shows the expansion in b).
b) A gas expands from 2.00 L to 5.00 L against a constant external pressure of 101.3 kPa, absorbing 500 J of heat. Calculate the work, then the change in internal energy.
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Answer
b)w=−304 J and ΔU=+196 J
b) ΔV=5.00−2.00=+3.00 L. The work is w=−PΔV=−101.3×3.00=−304 kPa·L, that is −304 J. The sign is negative because the gas PUSHES the atmosphere: it does work. Then ΔU=q+w=500+(−304)=+196 J.
A liquid evaporates until its vapour reaches an equilibrium pressure that depends only on temperature. When this pressure reaches the external pressure, boiling begins, throughout the volume and no longer only at the surface.
For water: ΔHvap=40.7 kJ/mol, boiling at 100 °C under 101.3 kPa. R = 8.314 J/(mol·K).
The diagram below shows the three situations.
d) In a pressure cooker, the pressure reaches 202.6 kPa. Calculate the new boiling point and explain the saving in cooking time.
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Answer
d)394.0 K, or 120.8 °C: about four times faster, through kinetics
d) With P2=202.6 kPa, the same calculation gives T2=394.0 K, or 120.8 °C. The pressure cooker therefore cooks at 121 °C instead of 100. The time saved comes from KINETICS, not thermodynamics: by the rule of thumb that reaction rates double for every 10 degrees, a rise of 21 degrees multiplies the rate of the cooking reactions by about four. It is also the standard sterilization temperature in an autoclave, chosen precisely because it destroys bacterial spores.
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