IB Chemistry SL Structure 1: Models of the particulate nature of matter questions

Answer: A. The close, regular arrangement before the change represents a solid; the widely separated particles after it represent a gas. The direction is solid → gas, which is sublimation.
B: reverses the direction: deposition is gas → solid.
C: identifies the initial solid but treats the final state as liquid; melting is solid → liquid.
D: identifies the final gas but assumes the initial state is liquid; vaporization is liquid → gas.
Remember: identify both states and the direction of change.

Answer: B. The abundance-weighted mean gives $63x+65(1-x)=63.60$ so the fraction of $^{63}\mathrm{Cu}$ is $x=0.70$ Both isotopes have 29 protons; their neutron numbers are $63-29=34$ and $65-29=36$ Each $2+$ ion has lost two electrons, leaving 27.

A: reverses the abundances, which would give a mean mass of 64.40.
C: treats the $2+$ charge as two extra electrons rather than two fewer.
D: reverses the subtraction for neutron number, assigning more neutrons to the lower-mass isotope.
Remember: Isotopes differ in neutron number; ion charge reflects electron gain or loss.

Answer: C. Each line corresponds to a photon emitted as an excited electron returns to a lower energy level. Violet light has a shorter wavelength and higher frequency than red light, so violet photons have greater energy.

A: confuses emission with absorption, which occurs as electrons move to higher energy levels.
B: identifies emission correctly but reverses the relative photon energies of red and violet light.
D: reverses the electron transition direction; emission occurs when electrons move to lower energy levels.

Remember: shorter wavelength means higher frequency and greater photon energy.

Answer: C. Each $\mathrm{O_2}$ molecule contains two oxygen atoms. Thus, $0.0400\times6.02\times10^{23}\times2=4.82\times10^{22}$ oxygen atoms.

A: uses $0.00400\ \text{mol}$ and counts $\mathrm{O_2}$ molecules rather than oxygen atoms.
B: counts $\mathrm{O_2}$ molecules but omits the two oxygen atoms in each molecule.
D: shifts the amount one place too high, using $0.400\ \text{mol}$

Remember: multiply by the number of specified entities in each molecule when counting atoms.

Answer: B. The molar volume is $V/n = RT/P$ Both $T$ and $P$ double, so $T/P$ and the molar volume remain unchanged; with the amount fixed, the volume also remains unchanged.
A: considers the pressure increase but ignores the effect of the doubled temperature.
C: considers the temperature increase but ignores the effect of the doubled pressure.
D: multiplies the temperature and pressure factors instead of using their ratio.
Remember: at fixed amount, volume changes in the same proportion as molar volume.