what volume of concentrated hcl should be used to prepare 500 ml of a 00 m hcl solution

HESI A2

HESI A2 Chemistry

1. How much concentrated HCl should be used to prepare 500 mL of a 0.100 M HCl solution?

A. 75 mL
B. 100 mL
C. 125 mL
D. 150 mL

Correct answer: B

Rationale: To prepare a 0.100 M HCl solution with a volume of 500 mL, you can use the formula C1V1 = C2V2, where C1 is the concentration of the concentrated HCl solution, V1 is the volume of concentrated HCl solution used, C2 is the desired concentration (0.100 M), and V2 is the final volume (500 mL). Rearranging the formula to solve for V1, you get V1 = (C2V2) / C1. Plugging in the values (0.100 M)(500 mL) / C1 = 100 mL, which means 100 mL of concentrated HCl should be used to prepare 500 mL of a 0.100 M HCl solution. Therefore, the correct answer is 100 mL. Choice A (75 mL), Choice C (125 mL), and Choice D (150 mL) are incorrect as they do not match the calculated volume needed to prepare the desired concentration of HCl solution.

2. What is the correct electron configuration for nitrogen?

A. 1s² 2s²
B. 1s² 2s² 2p²
C. 1s² 2s² 2p³
D. 1s² 2s² 2p⁴

Correct answer: C

Rationale: The electron configuration of nitrogen is determined by its atomic number, which is 7. Nitrogen has 7 electrons. Following the order of filling orbitals, the electron configuration for nitrogen is 1s² 2s² 2p³. This means the first energy level is filled with 2 electrons in the 1s orbital, the second energy level is filled with 2 electrons in the 2s orbital, and 3 electrons in the 2p orbital. Each orbital can hold a specific number of electrons, and nitrogen, with its 7 electrons, fits this configuration. Choice A is incorrect because it does not account for all the electrons in the nitrogen atom. Choice B is incorrect as it only represents 6 electrons, not the 7 electrons in nitrogen. Choice D is incorrect as it represents 8 electrons, which is not the correct electron configuration for nitrogen.

3. What type of bond is present in sodium chloride?

A. Covalent
B. Ionic
C. Metallic
D. Hydrogen

Correct answer: B

Rationale: Ionic bonds are found in sodium chloride. In an ionic bond, one atom donates an electron to another atom, resulting in the formation of positively and negatively charged ions that are held together by electrostatic forces of attraction. Sodium chloride is a classic example of an ionic compound, where sodium (Na) donates an electron to chlorine (Cl), forming Na+ and Cl- ions that are attracted to each other, creating a crystal lattice structure. Covalent bonds involve the sharing of electron pairs between atoms, which is not the case in sodium chloride. Metallic bonds occur in metals where electrons are delocalized and shared across a lattice, unlike the specific transfer seen in ionic bonds. Hydrogen bonds are a type of intermolecular force, not the primary bond type present in sodium chloride.

4. What charge do Group IA elements have?

A. +1
B. +2
C. +3
D. 0

Correct answer: A

Rationale: Group IA elements, also known as alkali metals, have a +1 charge. They readily lose one electron to achieve a stable electron configuration, forming ions with a single positive charge. This makes +1 the correct choice. Choices B, C, and D are incorrect because alkali metals in Group IA typically lose one electron, so they do not have a +2, +3, or 0 charge.

5. What term refers to the average of the masses of each of its isotopes as they occur in nature?

A. Atomic number
B. Mass number
C. Atomic mass
D. Neutron number

Correct answer: C

Rationale: The correct answer is atomic mass. Atomic mass is the weighted average of the masses of an element's isotopes. It takes into account the abundance of each isotope in nature to provide a more accurate representation of the element's overall mass. Choice A, atomic number, represents the number of protons in an atom. Choice B, mass number, refers to the total number of protons and neutrons in an atom's nucleus. Choice D, neutron number, specifically focuses on the count of neutrons in an atom's nucleus. These choices do not directly relate to the average mass of isotopes as asked in the question.