Unit 3 · Properties of Substances and MixturesPractice · 物质与混合物的性质

Boiling only overcomes the forces between molecules; it doesn’t break the O–H or S–H bonds inside them, so A’s reason is wrong. $\ce{H2S}$ has 18 electrons and $\ce{H2O}$ only 10, so $\ce{H2S}$ really does have stronger dispersion forces. But in $\ce{H2O}$ the H is bonded to the very electronegative O, so the molecules form hydrogen bonds, which are much stronger than the dipole–dipole and dispersion forces in $\ce{H2S}$. $\ce{H2S}$ is bent and polar, so D is false.

A hydrogen bond needs two things: an H bonded to N, O, or F on one molecule (the donor) and a lone pair on N, O, or F on the other (the acceptor). Dimethyl ether and acetone have O lone pairs, so they can accept, but they have no O–H, so they can’t hydrogen-bond to each other. Mixed with water, water’s H can hydrogen-bond to their O. $\ce{NH3}$ is both a donor and an acceptor. $\ce{CH4}$ is neither, so it doesn’t hydrogen-bond to water.

In a covalent network solid every atom is joined to the rest by covalent bonds, so melting means breaking many covalent bonds: the solid is hard and melts only at a very high temperature. It has no mobile ions or electrons, so it doesn’t conduct as a solid or a liquid (graphite is the exception). An ionic solid would conduct when melted, a metal conducts as a solid, and a molecular solid usually melts at a low temperature and is soft.

First convert the temperature to kelvins: T = 27.0 + 273.15 = 300.15 K.

M = m ÷ n = 1.24 g ÷ 0.0389 mol = 31.9 g/mol, consistent with $\ce{O2}$ (32.00 g/mol). The mass and pressure have 3 significant figures, so the answer has 3. Forgetting to convert to kelvins gives 2.87 g/mol.

Each gas spreads through the whole 5.00 L, and at constant temperature $P_1V_1 = P_2V_2$:

By Dalton’s law, the total pressure is 0.600 + 0.480 = 1.08 atm. Adding the starting pressures (2.30 atm) or averaging them (1.15 atm) is wrong.

At the same temperature both kinds of atoms have the same average kinetic energy, ½mv², so the heavier Kr moves more slowly on average: the most probable speed is about 497 m/s for Ne and about 244 m/s for Kr. The area under each curve stands for all the atoms, so Kr’s narrower distribution has a taller peak. A assumes the same temperature means the same speeds, B swaps the two gases, and D’s dashed curve is faster than Ne, so it would be an atom lighter than Ne.

Each $\ce{Na2SO4}$ releases 2 $\ce{Na+}$ ions, so $[\ce{Na2SO4}]$ = 0.300 ÷ 2 = 0.150 M.

Forgetting to divide by 2 gives 10.7 g, which would make $[\ce{Na+}]$ twice the target.

$\ce{K2SO4}$ dissociates in water into $\ce{K+}$ and $\ce{SO4^2-}$ ions in a 2 : 1 ratio. The bonds inside $\ce{SO4^2-}$ are covalent, so it stays together as one ion. The partially negative O of each water molecule faces the cation $\ce{K+}$, and the partially positive H atoms face the anion $\ce{SO4^2-}$. B shows undissociated units, C has a 1 : 1 ratio, so its charges don’t balance, and D has the water molecules turned the wrong way.

Both compounds have only dispersion forces. Pentane has 42 electrons and hexane 50, so pentane’s smaller electron cloud is less polarizable and its intermolecular forces are weaker. At the same temperature it has the higher vapor pressure and the lower boiling point, so the first liquid collected is mostly pentane. Boiling overcomes only the forces between molecules; no covalent bonds inside the molecules break (B is wrong). Having the same polarity doesn’t stop fractional distillation, which depends on different boiling points (D is wrong).

Percent crystallized = (solubility at 80 °C − solubility at 20 °C) ÷ solubility at 80 °C. $\ce{KNO3}$: (169 − 31.6) ÷ 169 ≈ 81.3%; $\ce{NH4Cl}$: (65.6 − 37.2) ÷ 65.6 ≈ 43.3%; $\ce{KCl}$: (51.1 − 34) ÷ 51.1 ≈ 33.5%; $\ce{NaCl}$: (38.4 − 36) ÷ 38.4 ≈ 6.25%. $\ce{KNO3}$’s solubility changes most with temperature, so it is best purified by cooling. NaCl’s solubility barely changes, so its crystals are usually obtained by evaporating the water instead.

Each photon must carry at least the work function: $E = \frac{hc}{\lambda} \ge \Phi$, so the longest wavelength is

That’s green light. Only shorter (higher-energy) wavelengths eject electrons; red light can’t, however bright it is.

Calculated concentration = absorbance of the unknown ÷ slope of the calibration curve, so ask whether each mistake changes the top or the bottom. A: fingerprints absorb and scatter light, so the absorbance reads high and the concentration comes out high. D: the standards are really more dilute than recorded, so plotting them at the recorded concentrations gives a slope that’s too small, and the concentration comes out high (about 1.05 times the true value). B: the unknown is diluted, so its absorbance is low: too low. C: the standards are really more concentrated, so the slope is too large: too low. E: half the path length gives half the absorbance: too low (0.5 times).