how many electrons are shared in a single covalent bond

HESI A2

HESI A2 Chemistry

1. How many electrons are shared in a single covalent bond?

A. 1
B. 2
C. 3
D. 4

Correct answer: B

Rationale: The correct answer is B: '2'. In a single covalent bond, two electrons are shared between two atoms. Each atom contributes one electron to form the bond, resulting in the sharing of a total of two electrons. Choice A is incorrect because a single covalent bond involves the sharing of two electrons, not one. Choices C and D are incorrect as they do not represent the correct number of electrons shared in a single covalent bond.

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. The molar mass of glucose is 180 g/mol. If an IV solution contains 5 g of glucose in 100 g of water, what is the molarity of the solution?

A. 0.28M
B. 1.8M
C. 2.8M
D. 18M

Correct answer: C

Rationale: To calculate the molarity of the solution, we first need to determine the moles of solute (glucose) and solvent (water) separately. The molar mass of glucose is 180 g/mol. First, calculate the moles of glucose: 5 g / 180 g/mol = 0.02778 mol of glucose. Next, calculate the moles of water: 100 g / 18 g/mol = 5.56 mol of water. Now, calculate the total moles in the solution: 0.02778 mol glucose + 5.56 mol water = 5.5878 mol. Finally, calculate the molarity: Molarity = moles of solute / liters of solution. Since the total mass of the solution is 100 g + 5 g = 105 g = 0.105 kg, which is equal to 0.105 L, the molarity is 5.5878 mol / 0.105 L = 53.22 M, which rounds to 2.8M. Therefore, the correct answer is 2.8M. Choices A, B, and D are incorrect because they do not reflect the accurate molarity calculation based on the moles of solute and volume of the solution.

4. Which of these intermolecular forces might represent attraction between atoms of a noble gas?

A. Dipole-dipole interaction
B. London dispersion force
C. Keesom interaction
D. Hydrogen bonding

Correct answer: B

Rationale: Noble gases are non-polar molecules without a permanent dipole moment. The only intermolecular force applicable to noble gases is the London dispersion force, also known as Van der Waals forces. This force is a temporary attractive force resulting from the formation of temporary dipoles in non-polar molecules. Dipole-dipole interactions, Keesom interactions, and hydrogen bonding involve significant dipoles or hydrogen atoms bonded to electronegative atoms, which do not apply to noble gases.

5. What are the 3 types of radiation in nuclear chemistry?

A. Alpha, Beta, Delta
B. Alpha, Beta, Gamma
C. Gamma, Beta, Delta
D. Delta, Beta, Gamma

Correct answer: B

Rationale: The correct answer is B: Alpha, Beta, Gamma. In nuclear chemistry, the 3 types of radiation are alpha, beta, and gamma radiation. Alpha radiation consists of helium nuclei, beta radiation involves electrons or positrons, and gamma radiation is electromagnetic radiation of high frequency. Choice A is incorrect because 'Delta' is not a type of radiation in nuclear chemistry. Choice C is incorrect as it does not list alpha radiation. Choice D is incorrect as it lists the types in the wrong order and includes 'Delta' instead of alpha radiation.