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​:

Electrons have negligible mass.

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.

Explanation​:

Isotopes with fewer neutrons will have a lower density

Explanation​:
The number of arrows denotes the number of electrons the atom has. The orbital diagram shown here has 8 arrows which tells you that there are 8 electrons.
Choice A is not correct because C has only 6 electrons.
Choice D is also not correct because F has 9 electrons.
Choice C is incorrect as N has 7 electrons.
Among the choices, oxygen in its neutral form has 8 electrons. Hence, choice B is the correct answer.

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:​
The student must understand that gases deviate from an ideal gas when the temperature is low and when the pressure is high.
At low temperatures, the particles move slower and thus are able to form more intermolecular forces.
At high pressure the particles are closer together, also allowing for more plentiful attractive forces.
In both of these cases, intermolecular forces of attraction cannot be assumed to be negligible and so the gas under those conditions is no longer considered an ideal gas.
This question is categorized as medium because the student needs to carefully consider the answer choices. A pressure of 1 $atm$ is the same as 100 kPa. Likewise, 298 $K$ is the same as 25∘C. A pressure of 2 $atm$, while higher, is not high enough to drastically increase attractive forces. A pressure of 200 $atm$ would be consistent with the pressure in a compressed gas cylinder when the gas is in liquid form, and is therefore not behaving as an ideal gas.

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

Explanation​:
The lines converge as energy levels increase (increased distance from the nucleus).
Higher energies result in higher frequencies.

Explanation:
The graph of P vs T shows a linear graph (straight line passing through the origin) that rises upward. This graph shows that P is directly proportional to the absolute temperature. As the absolute temperature increases, the pressure of the gas also increases.

Explanation:
The s orbital has a spherical shape. In terms of energy, s has the lowest, followed by p and d orbital. Thus, Choices A, B, and D are all incorrect.

Explanation:​
Electrons fill sublevels according to the Aufbau Principle shown below: 

Following this order, $2p<4s<3d<4p$

Explanation:​
The student must understand the concepts behind an ideal gas which assumes negligent volume and insignificant attractive forces between atoms or molecules.
As the molar mass of a chemical species increases, the attractive forces between particles also increase. This causes a deviation from the ideal gas model as the relative molecular mass increases.
This question is categorized as medium difficulty because the student must apply logic to the definition of an ideal gas.

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​:
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​:
One of the laws of gases states that, for a fixed amount of gas held at a constant temperature, the volume of a gas is inversely proportional to its pressure.

Explanation:
The molar mass of sodium phosphate is 163.94 g mol^−1. Divide the mass ($20.5g$) of sodium phosphate by the molar mass to calculate the number of moles of the compound. The rest of the choices are incorrect.

Explanation:​
The $\mathrm{Mn}$ 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​:
The student must correctly understand the process of paper chromatography. Components of a mixture can be successfully separated using paper chromatography only if the components are soluble in the mobile phase (the solvent). The components of the mixture can be liquids or solids when pure.
This question is categorized as medium because the student must understand that solubility in the mobile phase is essential for a successful separation using paper chromatography. Beyond that, there are no calculations or other facts to recall.