HESI A2
Chemistry Hesi A2
1. Aluminum (Al) has 13 protons in its nucleus. What is the number of electrons in an Al3+ ion?
- A. 16
- B. 13
- C. 10
- D. 3
Correct answer: C
Rationale: Aluminum (Al) has an atomic number of 13, which indicates it normally has 13 electrons to balance the 13 protons in its nucleus. When Al forms an Al3+ ion, it loses 3 electrons to achieve a stable electron configuration. Therefore, the Al3+ ion will have 13 - 3 = 10 electrons. Choice A (16) is incorrect as it doesn't take into account the charge of the Al3+ ion. Choice B (13) is incorrect because the Al3+ ion has lost electrons. Choice D (3) is incorrect as it doesn't reflect the total number of electrons lost by the Al atom to form the Al3+ ion.
2. On the periodic table, where are atoms with the largest atomic radius located?
- A. At the top of their group
- B. In the middle of their group
- C. At the bottom of their group
- D. Along the right-hand side
Correct answer: C
Rationale: Atoms with the largest atomic radius are located at the bottom of their group on the periodic table. This is because atomic radius increases down a group due to the addition of more energy levels or shells of electrons. As you move down a group, the outermost electrons are further away from the nucleus, leading to an increase in atomic radius. Choice A 'At the top of their group' is incorrect because atomic radius decreases going up within a group. Choice B 'In the middle of their group' is incorrect as the atomic radius generally increases as you go down a group, not in the middle. Choice D 'Along the right-hand side' is incorrect because atomic radius tends to decrease from left to right across a period on the periodic table due to increased nuclear charge and effective nuclear charge.
3. 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.
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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