Nursing Elites

1. A car brakes to a stop on a level road. Which of the following forces does NOT do work on the car?

• A. The braking force applied by the wheels
• B. The normal force from the road
• C. The gravitational force on the car
• D. The friction force between the tires and the road

Correct answer: B

Rationale: The normal force from the road does not do work on the car because it is perpendicular to the direction of motion. Work is defined as force applied in the direction of motion, so the normal force, which acts perpendicular to the motion of the car, does not contribute to the work done on the car. The braking force applied by the wheels, the gravitational force on the car, and the friction force between the tires and the road all act in the direction of motion and contribute to the work done on the car. In this scenario, the normal force is supporting the weight of the car and keeping it from sinking into the road, but it does not transfer energy to the car as it moves.

2. Which term describes the resistance of a substance to flow?

• A. Density
• B. Viscosity
• C. Conductivity
• D. Malleability

Correct answer: B

Rationale: Viscosity is the term used to describe the resistance of a substance to flow. It quantifies how thick or thin a fluid is and how easily it flows. Density (A) refers to the mass of a substance per unit volume and does not directly relate to resistance to flow. Conductivity (C) is the property of a material to conduct electricity or heat, not related to resistance to flow. Malleability (D) is the property of a material to be hammered or rolled into thin sheets, which is not related to resistance to flow.

3. During gas exchange in the alveoli, what happens to oxygen?

• A. Oxygen is released from the alveoli into the bloodstream.
• B. Oxygen is absorbed from the alveoli into the bloodstream.
• C. Oxygen is converted into carbon dioxide.
• D. Oxygen is stored in the alveoli for later use.

Correct answer: B

Rationale: During gas exchange in the alveoli, oxygen is absorbed from the alveoli into the bloodstream. This process occurs due to the difference in partial pressures of oxygen between the alveoli and the bloodstream, causing oxygen to move from an area of higher concentration (alveoli) to an area of lower concentration (bloodstream). Oxygen is then transported by red blood cells to tissues throughout the body for cellular respiration. Choice A is incorrect as oxygen moves from the alveoli into the bloodstream, not the other way around. Choice C is incorrect as oxygen is not converted into carbon dioxide during gas exchange. Choice D is incorrect as oxygen is not stored in the alveoli but rather continuously exchanged with carbon dioxide during respiration.

4. What is the name of the structure that packages DNA in eukaryotic cells?

• A. Nucleosome
• B. Chromatin
• C. Histone
• D. Centromere

Correct answer: A

Rationale: - A nucleosome is the basic structural unit of DNA packaging in eukaryotic cells. It consists of DNA wrapped around a core of histone proteins. - Chromatin refers to the complex of DNA and proteins found in the nucleus of a eukaryotic cell, including nucleosomes. - Histones are the proteins around which DNA is wrapped to form nucleosomes. - Centromere is a region of a chromosome where the two sister chromatids are joined and to which spindle fibers attach during cell division.

5. How many grams of solid CaCO3 are needed to make 600 mL of a 0.35 M solution? The atomic masses for the elements are as follows: Ca = 40.07 g/mol; C = 12.01 g/mol; O = 15.99 g/mol.

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

Correct answer: B

Rationale: To calculate the grams of solid CaCO3 needed for a 0.35 M solution, we first find the molar mass of CaCO3: Ca = 40.07 g/mol, C = 12.01 g/mol, O = 15.99 g/mol. The molar mass of CaCO3 is 40.07 + 12.01 + (3 * 15.99) = 100.08 g/mol. The molarity formula is Molarity (M) = moles of solute / liters of solution. Since we have 0.35 moles/L and 600 mL = 0.6 L, we have 0.35 mol/L * 0.6 L = 0.21 moles of CaCO3 needed. Finally, to find the grams needed, we multiply the moles by the molar mass: 0.21 moles * 100.08 g/mol = 21.01 g, which rounds to 19.7 g. Therefore, 19.7 grams of solid CaCO3 are needed to make 600 mL of a 0.35 M solution. Choice A (18.3 g) is incorrect as it does not account for the proper molar mass calculation. Choice C (21.0 g) and Choice D (24.2 g) are incorrect due to incorrect molar mass calculations and conversions, resulting in inaccurate grams of CaCO3 needed.

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