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:
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:
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:
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:

N = n x NA = (4.80)(6.02 $\times$ 10^$)$ = 2.89 $\times$ 10^24 molecules of glucose

Explanation:
The empirical formula shows the simplest whole-number ratio of elements in a compound. The empirical formula can sometimes be the same as the molecular formula. Choices A, B, and D are all false statements.

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:
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:
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 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.

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:
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:
According to Aufbau’s principle, the arrangement of subshell in increasing energy is 1s 2s 2p 3s 3p 4s 3d 4p. Therefore, Choices A and B are both incorrect. Choice C is incorrect because 2d does not exist.

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:
Seawater is a mixture of water and salts. Sucrose, distilled water, and sodium chloride are all examples of compounds.

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

Explanation:

Electrons have negligible mass.

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:
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.

Explanation:
For p, $l$ is equal to 1. The number of electrons in a given subshell can be calculated using the formula: 2(2 l+1)=2((2)(1)+1)=6 electrons.

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

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

Following this order, $2p<4s<3d<4p$