ATI TEAS 7
ATI TEAS Science Questions
1. What term describes the front of the body, for example, the kneecap?
- A. Posterior
- B. Anterior
- C. Superior
- D. Inferior
Correct answer: B
Rationale: The correct answer is 'B: Anterior.' In anatomy, 'anterior' refers to the front of the body, like the kneecap. On the other hand, 'posterior' describes the back of the body. Choices 'C: Superior' and 'D: Inferior' relate to structures being above or below a reference point, respectively, and are not used to describe the front or back of the body. Therefore, 'Anterior' is the most appropriate term given the context provided in the question.
2. What energy transformation occurs when a guitar string vibrates to produce sound?
- A. Mechanical energy to thermal energy
- B. Kinetic energy to potential energy
- C. Electrical energy to sound energy
- D. Potential energy to kinetic energy
Correct answer: D
Rationale: The correct answer is D. When a guitar string vibrates to produce sound, the energy transformation that occurs is from potential energy (stored energy in the string when it is stretched) to kinetic energy (energy of motion as the string vibrates back and forth). As the string vibrates, its kinetic energy is transferred to the surrounding air molecules, producing sound energy. Choices A, B, and C are incorrect. Choice A, mechanical energy to thermal energy, does not align with the energy transformation involved in producing sound from a vibrating guitar string. Choice B, kinetic energy to potential energy, is the opposite of what happens when a guitar string vibrates. Choice C, electrical energy to sound energy, is not relevant to the energy conversion process in this scenario.
3. Which of the following is an example of an aromatic hydrocarbon?
- A. Ethane
- B. Benzene
- C. Propane
- D. Butene
Correct answer: B
Rationale: Benzene is indeed an example of an aromatic hydrocarbon. Aromatic hydrocarbons are characterized by having a cyclic structure with alternating single and double bonds (pi bonds). Benzene fits this description, making it aromatic. On the other hand, ethane, propane, and butene are aliphatic hydrocarbons, which do not have the distinct cyclic structure of aromatic hydrocarbons. Ethane, propane, and butene are aliphatic hydrocarbons, which contain only single bonds and are not cyclic in nature. Therefore, they are not examples of aromatic hydrocarbons.
4. How many moles of water are produced when 0.5 moles of methane (CH4) react with excess oxygen?
- A. 0.5 moles
- B. 1 mole
- C. 2 moles
- D. 3 moles
Correct answer: B
Rationale: The balanced chemical equation for the combustion of methane is: CH4 + 2O2 -> CO2 + 2H2O. This equation shows that 1 mole of methane produces 2 moles of water. Therefore, when 0.5 moles of methane react, they will produce 1 mole of water. The ratio of water to methane is 2:1, meaning that for every mole of methane that reacts, it produces 2 moles of water. Since only 0.5 moles of methane are reacting, the amount of water produced will be half of what would be produced if 1 mole of methane reacted, resulting in 1 mole of water being produced. Choice A is incorrect because it does not consider the stoichiometry of the reaction. Choices C and D are incorrect as they do not align with the balanced chemical equation and the stoichiometric ratios involved in the reaction.
5. Which of the following units is used to express concentration as a mass of solute per unit volume of solution?
- A. Molality (m)
- B. Molarity (M)
- C. Weight percent (%)
- D. Parts per million (ppm)
Correct answer: A
Rationale: Molality (m) is the unit used to express concentration as a mass of solute per unit volume of solution. It is calculated by dividing the mass of solute in grams by the mass of the solvent in kilograms. Molality is preferred over molarity when there are large temperature variations as it is temperature-independent, making it a more accurate measure of concentration. Molarity (B) is the unit used to express concentration as moles of solute per liter of solution, weight percent (C) is the mass of solute per 100 parts of the total mass of the solution, and parts per million (D) is used to express very small concentrations where 1 ppm is equivalent to 1 mg of solute per liter of solution.
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