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

Explanation:
Vanadium has an atomic number of 23 therefore statement I is true. It also has a mass number of 51, therefore statement II is also true as nucleons include both protons and neutrons. Statement III is false as the closest noble gas is Ar, not Ne. The correct condensed electron configuration would be [Ar]4s23d^3. Therefore, the correct answer is A.

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:
There is a large increase between the 3rd and 4th ionization energies which shows that the element is in group 13.
Large increases in ionization energy indicate electrons being removed from a shell closer to the nucleus.

Explanation:
The shortest wavelength corresponds to a higher frequency and therefore a higher energy transition. The 5p to $1s$ represents the largest energy drop and therefore would produce the shortest wavelength.

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

Following this order, 2p<4s<3d<4p

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:
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:
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 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.

Explanation:

Isotopes are atoms of the same element with different numbers of neutrons that differ only in terms of their physical properties.

Explanation:

Isotopes with fewer neutrons will have a lower density

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:
Second ionization energies are higher than the first as the electrons are being removed from a positive ion resulting in a stronger electrostatic attraction.
First ionization energies do increase across a period due to increasing nuclear charge.
First ionization energies also decrease down a group due to increased shielding (number of shells).