Explanation:
Liquids have an indefinite shape (take the shape of the container) but have a definite volume. Choice A describes solids. Choices B and D describe the characteristics of gases.

Explanation​:
Statement I is true because emission spectra are produced when photons are released from excited electrons returning to lower energy levels.
Statement II is true because this is Pauli’s Exclusion principle.
Statement III is also true because it is included in Pauli’s Exclusion principle.

Explanation​:
Choice B is incorrect because electrons are not located in the nucleus. Choice C is incorrect because an orbital diagram is just a representation of electrons in an orbital. Choice D is incorrect because electrons can be found in orbit (a specific energy level around the positively charged nucleus) but the probability is relatively low. Choice A is the correct answer: orbitals are areas/ regions in space where the probability of finding an electron is very high.

Explanation​:
Choice A is incorrect because the orbital diagram does not satisfy Hund's rule wherein all orbitals must be singly occupied before pairing.
Choice B is incorrect because Pauli's Exclusion principle is not obeyed by having two electrons with the same spin in one orbital.
Choice C is correct because, according to Hund's rule, the orbitals must be half-filled first before pairing, and the paired orbital must have electrons in an opposite spin.
Choice D is incorrect because the 2s orbital must be filled first before filling the 2p orbital, as illustrated by Aufbau's Principle.

Explanation​:
The student must first understand that each component of the mixture will result in one spot on the paper chromatogram. If the recrystallization was successful, then the purified dye should only have one dot.
Therefore, answer choice B is incorrect since the sample contains more than one component after recrystallization.
Answer choice C indicates that the dye is no longer in the sample, and answer choice D indicates the contaminant was not successfully separated from the dye. So the paper chromatogram is unable to identify any impurities that are still present.
In answer choice A, before recrystallization, shows two components (the dye and contaminant), and after, there is only one component showing successful recrystallization.
This question is categorized as medium since the student must understand the premise and purpose of paper chromatography and apply it to experimental results

Explanation:
To answer this question, the student must understand the properties of the mixture's components.
The wood chips would not dissolve in the water, so they are filtered out first. This will leave the sodium chloride dissolved in the water. The water can be evaporated to remove the water (to isolate the salt).
Paper chromatography would be ineffective in this scenario as it is used for separating substances such as dyes, pigments, and proteins.
This question is categorized as medium complexity because the student must understand the properties of matter and separation techniques.

Explanation:
Ions do not have the same numbers of protons and electrons. Only species B and C do not have the same number of protons and electrons, so the correct answer is B.

Explanation:​
Condensation (gas to liquid) requires making bonds and therefore, releases energy. Sublimation (solid to gas) requires breaking bonds and therefore absorbs energy.

Explanation​:
At high pressures and very low temperatures, the inter-particle forces strengthen, spaces between particles decrease causing the deviation from ideal gas behaviour PV/nRT​ > 1.

Explanation​:
Relative atomic mass of Br=(0.5069)(78.92)+(0.4931)(80.92)=79. The rest of the choices are incorrect relative atomic masses of bromine.

Alternative Method:

79Br atomic mass can be rounded up to 79

81Br atomic mass can be rounded up to 81

Since the relative abundance of both isotopes is about 50% each, the relative atomic mass of Bromine should be close to 80. Choice D corresponds to that estimation.

Explanation​:
The electron configuration for S is 1s2 2s2 2p6 3s2 3p4, whereas for $\mathrm{P}$, it is 1s2 2s2 2p6 3s2 3p3.
Due to the increased stability of P’s half-filled p-shell, more energy is required to remove an electron from a $\mathrm{P}$ atom than an $\mathrm{S}$ atom. The electron-electron repulsion in Sulfur makes it easier to remove the first electron.
First ionization energy increases across a period (due to a decrease in atomic radii and an increase in nuclear charge). In period 3, Na is the first element, and Ar is the last to the right.

The atomic number is the number of protons. The mass number is the sum of the number of protons and neutrons. For a neutral atom, the number of electrons is equal to the number of protons.

Explanation​:
Successive ionization energies are larger, as electrons are removed, there is less electron-electron repulsion, so the remaining electrons are pulled closer to the nucleus, and more energy is required to remove them.

Explanation​:
There is an inverse relationship between frequency and wavelength. If the wavelength is doubled, the frequency will decrease by half. Therefore, Choices A, C, and D are all incorrect.

Explanation:
For p, $l$ is equal to 1. The number of electrons in a given subshell can be calculated using the formula: $6$ electrons.

The p subshell holds 3 orbitals. Each orbital can fit 2 electrons. Therefore, p subshell holds 6 electrons.

The mass number is the sum of the number of protons and neutrons. For a neutral atom, the number of electrons is equal to the number of protons

Explanation:
Isotopes have the same number of protons but different numbers of neutrons.
Species A and B have the same number of protons but also the same number of neutrons so cannot be isotopes.
Species C is the only species in the table with 9 protons so cannot have an isotope with another species in the table.
This leaves species D and E which both have the same number of protons but a different number of neutrons. Therefore, the correct answer is D.

Explanation:​
Electron configuration of Al^+: [Ne] 3s^2
Electron configuration of Zn: [Ar] 4s^2 3d^10
Electron configuration of Ca: [Ar] 4s2

Electron configuration of Sc: [Ar] 4s^2 3d^1.
Sc will lose 2 electrons in the 4s sublevel and 1 from the 3d sublevel, leaving a complete 3rd energy level with 8 valence electrons in Sc^3+ (1s^2 2s^2 2p^6 3s^2 3p^6).

Explanation:​
The $\mathrm{n}$ atom and Mn^2+ ion have the following orbital diagrams:

The valence electrons and the two lost in the 2+ ion are the $4s$ electrons so the unpaired electrons remain 5.

Explanation​:
Choice A is the correct answer because it satisfies Aufbau's principle and the number of electrons in the electron configuration aligns with the same number of electrons in the orbital diagram.

Explanation:
The student must first recall that the component with the greatest affinity for the mobile phase will travel the furthest in paper chromatography. In this case, spot #2 is the furthest from the starting line, so it has the greatest affinity for the mobile phase.
In paper chromatography, the mobile phase is always the solvent (water, in this case), while the stationary phase is the paper.
This question is categorized as easy because the student must only recall basic definitions and concepts for paper chromatography.

Explanation:
1 mole is the number of H2 molecules in 2.02 g of H2. Choices A, B, and C all conform to the definition of a mole