Nursing Elites

1. What is the maximum volume of air that the lungs can hold after a full forced inhalation?

• A. Inspiratory capacity
• B. Tidal volume
• C. Total lung capacity
• D. Vital capacity

Correct answer: C

Rationale: Total lung capacity is the correct term for the maximum volume of air that the lungs can hold after a full forced inhalation. It represents the sum of all lung volumes, including tidal volume, inspiratory reserve volume, and expiratory reserve volume. Inspiratory capacity refers to the maximum volume of air inspired from the end-expiratory level. Tidal volume is the volume of air inspired or expired during normal breathing. Vital capacity is the maximum volume of air that can be exhaled after a maximum inhalation, not the total volume the lungs can hold.

2. What is the relationship between the Pauli exclusion principle and the structure of the atom?

• A. It defines the maximum number of electrons allowed in each energy level.
• B. It explains why oppositely charged particles attract each other.
• C. It describes the wave-particle duality of electrons.
• D. It determines the arrangement of protons and neutrons in the nucleus.

Correct answer: A

Rationale: The Pauli exclusion principle states that no two electrons in an atom can have the same set of quantum numbers. This principle defines the maximum number of electrons allowed in each energy level, influencing the structure of the atom. Choice B is incorrect as it refers to the concept of electrostatic attraction, not directly related to the Pauli exclusion principle. Choice C is incorrect as it pertains to the wave-particle duality, a different aspect of quantum mechanics. Choice D is incorrect as it relates to the arrangement of protons and neutrons in the nucleus, not governed by the Pauli exclusion principle.

3. Which of the following is responsible for generating the electrical impulses that regulate the heartbeat?

• A. Aorta
• B. Pulmonary veins
• C. Coronary arteries
• D. Sinoatrial (SA) node

Correct answer: D

Rationale: The Sinoatrial (SA) node is responsible for generating the electrical impulses that regulate the heartbeat. It is often referred to as the heart's natural pacemaker because it initiates the electrical signals that coordinate the heart's contractions. The other options listed (Aorta, Pulmonary veins, Coronary arteries) do not play a direct role in generating the electrical impulses for the heartbeat. The Aorta is a large artery that carries oxygen-rich blood away from the heart, the Pulmonary veins carry oxygen-rich blood from the lungs to the heart, and the Coronary arteries supply blood to the heart muscle itself, but none of them are involved in generating the electrical impulses that regulate the heartbeat.

4. The kidneys are bean-shaped organs responsible for filtering waste products from the blood. What is the main nitrogenous waste product the kidneys eliminate?

• A. Carbon dioxide
• B. Ammonia
• C. Urea
• D. Glucose

Correct answer: C

Rationale: The correct answer is C, urea. Urea is the main nitrogenous waste product eliminated by the kidneys. It is produced in the liver from protein metabolism and excreted in urine. Carbon dioxide is eliminated through the lungs as a waste product of cellular respiration and not by the kidneys. Ammonia, a toxic waste product, is converted to urea in the liver before being excreted by the kidneys. Glucose is a sugar that is reabsorbed by the kidneys and not excreted as waste.

5. How many grams of solid CaCO3 are needed to make 600 mL of a 35 M solution? The atomic masses for the elements are as follows: Ca = 40.1 g/mol; C = 12.01 g/mol; O = 16.00 g/mol.

• A. 18.3 g
• B. 19.7 g
• C. 21.0 g
• D. 24.2 g

Correct answer: B

Rationale: 1. First, calculate the molar mass of CaCO3 by adding the atomic masses of Ca, C, and 3 O atoms: 40.1 + 12.01 + (3 * 16.00) = 100.13 g/mol. 2. Calculate the number of moles in 600 mL of a 35 M solution: 600 mL * 35 mol/L = 21,000 mmol. 3. Convert moles to grams using the molar mass of CaCO3: 21,000 mmol * (100.13 g/mol / 1000 mmol/mol) = 2,102.73 g. 4. Therefore, you would need 19.7 g of solid CaCO3 to make 600 mL of a 35 M solution.

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