Nursing Elites

1. What are the products of combustion of a hydrocarbon in excess oxygen?

• A. Carbon dioxide and water
• B. Naphthalene
• C. Chlorine and bromine
• D. Carbonium ions

Correct answer: A

Rationale: The correct answer is A: Carbon dioxide and water. During the combustion of a hydrocarbon in excess oxygen, the hydrocarbon reacts to produce carbon dioxide and water vapor as the final products. This reaction is known as complete combustion, where the hydrocarbon combines with oxygen to form carbon dioxide and water. Choices B, C, and D are incorrect because naphthalene is a specific hydrocarbon compound, chlorine and bromine are not typically formed during the combustion of hydrocarbons in excess oxygen, and carbonium ions are not the products of this reaction.

2. Which substance shows a decrease in solubility in water with an increase in temperature?

• A. NaCl
• B. O
• C. KI
• D. CaCl

Correct answer: C

Rationale: Potassium iodide (KI) shows a decrease in solubility in water with an increase in temperature. This is due to the dissolution of KI in water being an endothermic process. When the temperature rises, the equilibrium shifts toward the solid state, leading to a decrease in solubility. Therefore, as the temperature increases, KI becomes less soluble in water. Choice A (NaCl) and Choice D (CaCl) do not exhibit a decrease in solubility with an increase in temperature. NaCl and CaCl are generally more soluble in water at higher temperatures. Choice B (Oxygen) is a gas and not typically considered in solubility discussions involving solids or liquids dissolving in water.

3. What is the molarity of a solution containing 45 moles of NaCl in 4 liters?

• A. 0.11 M NaCl
• B. 0.45 M NaCl
• C. 1.8 M NaCl
• D. 8.9 M NaCl

Correct answer: A

Rationale: To calculate the molarity of a solution, you use the formula: Molarity (M) = moles of solute / liters of solution. In this case, M = 45 moles / 4 L = 11.25 M. The correct answer is 0.11 M NaCl. Choice B is incorrect as it doesn't match the calculated value. Choice C is also incorrect as it is significantly higher than the correct molarity. Choice D is incorrect as it is excessively high compared to the calculated value.

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 type of starch is glycogen?

• A. Plant starch
• B. Animal starch
• C. Glucose
• D. Cellulose

Correct answer: B

Rationale: Glycogen is classified as animal starch, not plant starch. It is the storage form of glucose in animals, primarily found in the liver and muscles. Choice A (Plant starch) is incorrect because glycogen is not derived from plants. Choice C (Glucose) is incorrect as glucose is a monosaccharide and not a type of starch. Choice D (Cellulose) is incorrect as cellulose is a structural polysaccharide found in plant cell walls, not the same as glycogen.

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