HESI A2 Chemistry

HESI A2 Chemistry Related

Review HESI A2 Chemistry related questions and content

Which two elements are most alike in reactivity?

  • A. He and H
  • B. K and Ar
  • C. Cl and P
  • D. Ba and Mg
Correct Answer: C

Rationale: Chlorine (Cl) and Phosphorus (P) are most alike in reactivity among the given pairs. Both elements are nonmetals and belong to Group 7 (halogens) and Group 15 (nitrogen group), respectively. They have similar electronic configurations and can form compounds by gaining or sharing electrons. Chlorine is highly reactive and can easily form ionic compounds, while phosphorus also shows a range of reactivity in its compounds. Choice A (He and H) is incorrect because helium (He) is a noble gas and hydrogen (H) is a nonmetal, so they are not similar in reactivity. Choice B (K and Ar) is incorrect as potassium (K) is a metal and argon (Ar) is a noble gas, having different reactivities. Choice D (Ba and Mg) is incorrect because barium (Ba) and magnesium (Mg) are both metals, but their reactivities differ due to their positions in the periodic table.

You may also like to solve these questions

Which of these intermolecular forces would result in the lowest boiling point?

  • A. Dipole-dipole interaction
  • B. London dispersion force
  • C. Keesom interaction
  • D. Hydrogen bonding
Correct Answer: B

Rationale: The London dispersion force is the weakest intermolecular force among the options provided. These forces are present in all molecules and are caused by temporary fluctuations in electron density, resulting in temporary dipoles. Since London dispersion forces are generally weaker than dipole-dipole interactions, Keesom interactions, and hydrogen bonding, a substance with London dispersion forces as the primary intermolecular force would have the lowest boiling point due to the weaker intermolecular forces holding the molecules together. Dipole-dipole interactions, Keesom interactions, and hydrogen bonding are stronger intermolecular forces compared to London dispersion forces, resulting in higher boiling points for substances that exhibit these interactions.

If fifty-six kilograms of a radioactive substance has a half-life of 12 days, how many days will it take the substance to decay naturally to only 7 kilograms?

  • A. 8
  • B. 12
  • C. 36
  • D. 48
Correct Answer: C

Rationale: To decay from 56 kg to 7 kg, the substance needs to go through 3 half-lives (56 kg · 2 · 2 · 2 = 7 kg). Since each half-life is 12 days, the total time required is 12 days per half-life x 3 half-lives = 36 days. Choice A is incorrect because it does not consider the concept of half-lives. Choice B is incorrect because it represents the duration of a single half-life, not the total time required for the decay. Choice D is incorrect as it does not account for the multiple half-lives needed for the substance to decay from 56 kg to 7 kg.

What is a pentose sugar?

  • A. A 6 carbon sugar
  • B. A 4 carbon sugar
  • C. A 5 carbon sugar
  • D. A 7 carbon sugar
Correct Answer: C

Rationale: A pentose is a 5-carbon sugar, as the prefix 'pent-' signifies five. Therefore, a pentose sugar consists of five carbon atoms. Choice A, 'A 6 carbon sugar,' is incorrect as a pentose sugar specifically has five carbons. Choice B, 'A 4 carbon sugar,' is incorrect as it describes a tetrose sugar, which has four carbons. Choice D, 'A 7 carbon sugar,' is incorrect as it does not correspond to the definition of a pentose sugar.

What is the name of the compound CH₃-CH₂-CH₂-CH₃?

  • A. Cyclobutane
  • B. Butane
  • C. Butene
  • D. Butyne
Correct Answer: B

Rationale: The compound CH₃-CH₂-CH₂-CH₃ is named butane. Butane is a straight-chain alkane comprising four carbon atoms connected by single bonds. The prefix 'but-' denotes the presence of four carbon atoms, while the suffix '-ane' indicates it is an alkane with single bonds between the carbon atoms. Choice A, Cyclobutane, is incorrect as it refers to a cyclic hydrocarbon with four carbon atoms in a ring structure. Choice C, Butene, is incorrect because it is an alkene with a double bond between two carbon atoms, not a saturated hydrocarbon like butane. Choice D, Butyne, is also incorrect as it is an alkyne with a triple bond between two carbon atoms, unlike the single bonds in butane.

You contain two odorous gases in vials with porous plugs. Gas A has twice the mass of Gas B. Which observation is most likely?

  • A. You will smell Gas A before you smell Gas B.
  • B. You will smell Gas B before you smell Gas A.
  • C. You will smell Gas A but not Gas B.
  • D. You will smell Gas B but not Gas A.
Correct Answer: A

Rationale: The rate of effusion of a gas is inversely proportional to the square root of its molar mass. Since Gas A has twice the mass of Gas B, Gas A will effuse more slowly than Gas B. Therefore, you will likely smell Gas A before you smell Gas B as Gas A will escape and diffuse through the porous plug at a slower rate compared to Gas B. Choice A is correct because Gas A, with its higher molar mass, will take longer to effuse through the porous plug, causing you to smell it first. Choices B, C, and D are incorrect as they do not consider the relationship between molar mass and effusion rate.

Copyright © 2025 Nokea. All rights reserved.