Nursing Elites

1. What is the difference between type 1 and type 2 diabetes, both related to insulin and blood sugar?

• A. Type 1 is caused by insulin resistance, type 2 by insulin deficiency.
• B. Type 1 is temporary, type 2 is permanent.
• C. Type 1 affects adults, type 2 affects children.
• D. Type 1 is treated with diet only, type 2 requires medication.

Correct answer: A

Rationale: The correct difference between type 1 and type 2 diabetes lies in their underlying causes. Type 1 diabetes is caused by the body's inability to produce insulin, while type 2 diabetes is characterized by insulin resistance, where the body doesn't effectively use the insulin it produces. Insulin resistance is a key feature of type 2 diabetes, distinguishing it from type 1 where insulin deficiency is the primary issue. Choices B, C, and D are incorrect. Type 1 and type 2 diabetes can both be chronic conditions, affecting individuals of different age groups, and typically require a combination of diet, exercise, and medication for management.

2. Which type of blood vessel carries oxygenated blood away from the heart?

• A. Vein
• B. Artery
• C. Capillary
• D. Lymphatic vessel

Correct answer: B

Rationale: The correct answer is B. Arteries carry oxygenated blood away from the heart to the rest of the body, delivering nutrients and oxygen to the tissues. Veins, on the other hand, carry deoxygenated blood back to the heart. Capillaries are where the exchange of gases, nutrients, and waste products occurs between the blood and tissues. Lymphatic vessels are responsible for transporting lymph, which is a clear fluid containing white blood cells and waste products, and play a key role in the immune system.

3. Long bones are one of the five major types of bones in the human body. All of the following bones are long bones, EXCEPT

• A. Thighs
• B. Forearms
• C. Ankles
• D. Fingers

Correct answer: D

Rationale: Long bones are one of the five major types of bones in the human body, characterized by their elongated shape. Thighs, forearms, and ankles are examples of long bones as they are longer than they are wide and have a tubular structure, aiding in support and movement. Fingers, however, are categorized as short bones due to their small size and shape. Short bones like fingers are essential for providing dexterity and precise movements rather than supporting weight or acting as levers for movement. Therefore, the correct answer is D, Fingers, as they are not classified as long bones.

4. What type of tissue is bone composed of?

• A. Epithelial tissue
• B. Connective tissue
• C. Hard connective tissue
• D. Muscle tissue

Correct answer: B

Rationale: Bone is composed of connective tissue. Connective tissues are characterized by having cells scattered within an extracellular matrix. In the case of bone, the extracellular matrix is mineralized, giving bone its hardness and strength. Choice A, epithelial tissue, is not correct as bone is not primarily composed of epithelial cells. Choice C, hard connective tissue, is not a recognized category of tissue in the scientific classification; bone is classified under connective tissue. Choice D, muscle tissue, is incorrect as bone and muscle tissues are distinct types of tissues with different structures and functions.

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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