HESI A2
Chemistry Hesi A2
1. Which of the following compounds is ionic?
- A. NaCl
- B. H₂O
- C. HCl
- D. NH₃
Correct answer: A
Rationale: The correct answer is NaCl (sodium chloride). Ionic compounds are formed by the transfer of electrons between a metal and a nonmetal. In NaCl, sodium (Na) is a metal, and chlorine (Cl) is a nonmetal. Sodium donates an electron to chlorine, leading to the formation of the ionic bond between them. This results in the formation of an ionic compound, where positively charged sodium ions are attracted to negatively charged chloride ions, creating a crystal lattice structure. Choices B, C, and D are not ionic compounds. H₂O (water) is a covalent compound formed by the sharing of electrons between two nonmetals (oxygen and hydrogen). HCl (hydrogen chloride) and NH₃ (ammonia) are also covalent compounds involving nonmetals sharing electrons, not transferring them.
2. 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.
3. A radioactive isotope has a half-life of 20 years. How many grams of a 6-gram sample will remain after 40 years?
- A. 8
- B. 6
- C. 3
- D. 1.5
Correct answer: C
Rationale: The half-life of a radioactive isotope is the time it takes for half of the original sample to decay. After each half-life period, half of the initial sample remains. In this case, after the first 20 years, half of the 6-gram sample (3 grams) will remain. After another 20 years (total of 40 years), half of the remaining 3 grams will remain, which is 1.5 grams. Therefore, 3 grams will be left after 40 years. Choice A is incorrect as it doesn't consider the concept of half-life and incorrectly suggests an increase in the sample. Choice B is incorrect as it assumes no decay over time. Choice D is incorrect as it miscalculates the remaining amount after two half-life periods.
4. In the solid state, you would expect a nonmetal to be _________.
- A. brittle
- B. lustrous
- C. malleable
- D. conductive
Correct answer: A
Rationale: In the solid state, you would expect a nonmetal to be brittle. Nonmetals generally lack the malleability and ductility of metals, which makes them prone to being brittle and easily fractured under stress. This property is due to the lack of metallic bonding in nonmetals, which results in a more rigid and less flexible structure. Choice B, 'lustrous,' is incorrect because nonmetals typically do not exhibit a shiny or reflective surface like metals do. Choice C, 'malleable,' is also incorrect as nonmetals lack the ability to be hammered or rolled into thin sheets like metals. Choice D, 'conductive,' is incorrect since nonmetals are generally poor conductors of electricity compared to metals.
5. A chemist takes 100 mL of a 40 g NaCl solution and dilutes it to 1L. What is the concentration (molarity) of the new solution?
- A. 0.04 M NaCl
- B. 0.25 M NaCl
- C. 0.40 M NaCl
- D. 2.5 M NaCl
Correct answer: C
Rationale: Initially, the chemist has 40 g of NaCl in 100 mL of solution. To find the initial molarity, we need to calculate the number of moles of NaCl using the molar mass of NaCl (58.44 g/mol). After dilution to 1 L, the molarity of the new solution can be calculated by dividing the moles of NaCl by the total volume in liters. Therefore, the concentration (molarity) of the new solution is 0.40 M NaCl. Choice A (0.04 M NaCl) is incorrect because it doesn't consider the correct molar concentration after dilution. Choice B (0.25 M NaCl) is incorrect as it also doesn't account for the correct molar concentration post-dilution. Choice D (2.5 M NaCl) is incorrect as it is too concentrated given the initial amount of NaCl and the dilution factor.
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