UNE CHEM 1011 MIDTERM EXAM STUDY GUIDE
ACTUAL EXAM QUESTIONS AND VERIFIED
ANSWERS (2026 EDITION)
This comprehensive exam study guide provides actual practice
questions and verified answers meticulously tailored for the UNE
CHEM 1011 Midterm. Each question features structured multiple-
choice options with italicized answers and deep, bolded rationales that
dissect core chemistry principles. It serves as an optimized, high-yield
asset engineered to maximize test scores and streamline complex
retention for ultimate student success.
1. Which of the following represents a pair of isotopes?
A) \(^{14}\text{N}\) and \(^{14}\text{O}\)
B) \(^{12}\text{C}\) and \(^{14}\text{C}\)
C) \(\text{O}_{2}\) and \(\text{O}_{3}\)
D) \(\text{Na}^{+}\) and \(\text{Na}\)
Answer: B
Rationale: Isotopes are atoms of the same element with the
same number of protons (same atomic number) but
different numbers of neutrons (different mass numbers).
Carbon-12 and Carbon-14 both have 6 protons but differ in
neutrons.
2. What is the empirical formula of a compound that is
40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass?
A) \(\text{CHO}\)
B) \(\text{CH}_2\text{O}\)
,C) \(\text{C}_2\text{H}_4\text{O}_2\)
D) \(\text{CHO}_{2}\)
Answer: B
Rationale: Assuming a 100 g sample gives 40.0 g C, 6.7 g H,
and 53.3 g O. Converting to moles: \(40..01 = 3.33\text{
mol C}\), \(6..008 = 6.65\text{ mol H}\), and \(53..00
= 3.33\text{ mol O}\). Dividing by the smallest value (3.33)
yields a 1:2:1 ratio, resulting in \(\text{CH}_2\text{O}\).
3. How many protons, neutrons, and electrons are in a
\(^{37}\text{Cl}^-\) ion?
A) 17 protons, 20 neutrons, 17 electrons
B) 17 protons, 20 neutrons, 18 electrons
C) 18 protons, 19 neutrons, 17 electrons
D) 17 protons, 17 neutrons, 18 electrons
Answer: B
Rationale: Chlorine always has an atomic number of 17 (17
protons). The mass number is 37, so neutrons = \(37 - 17 =
20\). Because it has a -1 charge, it has gained one electron,
totaling 18 electrons.
4. Which element has the ground-state electron
configuration \([Ar] 4s^2 3d^{10} 4p^3\)?
A) \(\text{V}\)
B) \(\text{As}\)
C) \(\text{Se}\)
,D) \(\text{Sb}\)
Answer: B
Rationale: Counting the total electrons past Argon (18): \(18
+ 2 + 10 + 3 = 33\). The element with atomic number 33 on
the periodic table is Arsenic (\(\text{As}\)).
5. What is the molarity of a solution made by dissolving 4.0
grams of \(\text{NaOH}\) (molar mass = 40.0 g/mol) in
enough water to make 250 mL of solution?
A) 0.10 M
B) 0.40 M
C) 1.0 M
D) 0.25 M
Answer: B
Rationale: Moles of \(\text{NaOH} = 4.0\text{ g} / 40.0\text{
g/mol} = 0.10\text{ mol}\). Volume in liters = \(0.250\text{ L}\).
Molarity = \(0.10\text{ mol} / 0.250\text{ L} = 0.40\text{ M}\).
6. Identify the limiting reactant when 2.0 moles of
\(\text{H}_{2}\) react with 2.0 moles of \(\text{O}_{2}\) to form
water via the equation: \(2\text{H}_2(g) + \text{O}_2(g)
\rightarrow 2\text{H}_2\text{O}(g)\).
A) \(\text{H}_{2}\)
B) \(\text{O}_{2}\)
C) \(\text{H}_2\text{O}\)
D) Neither reactant is limiting
, Answer: A
Rationale: According to the stoichiometry, 2.0 moles of
\(\text{H}_{2}\) require \(1.0\text{ mole}\) of \(\text{O}_{2}\).
Since we have 2.0 moles of \(\text{O}_{2}\), \(\text{O}_{2}\) is
in excess and \(\text{H}_{2}\) will run out first, making it the
limiting reactant.
7. Which of the following molecules has a tetrahedral
molecular geometry?
A) \(\text{BF}_{3}\)
B) \(\text{NH}_{3}\)
C) \(\text{CH}_{4}\)
D) \(\text{SF}_{4}\)
Answer: C
Rationale: Methane (\(\text{CH}_{4}\)) has 4 bonding pairs
and 0 lone pairs around the central carbon atom, resulting
in a perfect tetrahedral geometry. \(\text{NH}_{3}\) is
trigonal pyramidal due to a lone pair.
8. What is the formal charge on the central nitrogen atom in
the nitrate ion (\(\text{NO}_{3}^{-}\))?
A) -1
B) 0
C) +1
D) +2
Answer: C
ACTUAL EXAM QUESTIONS AND VERIFIED
ANSWERS (2026 EDITION)
This comprehensive exam study guide provides actual practice
questions and verified answers meticulously tailored for the UNE
CHEM 1011 Midterm. Each question features structured multiple-
choice options with italicized answers and deep, bolded rationales that
dissect core chemistry principles. It serves as an optimized, high-yield
asset engineered to maximize test scores and streamline complex
retention for ultimate student success.
1. Which of the following represents a pair of isotopes?
A) \(^{14}\text{N}\) and \(^{14}\text{O}\)
B) \(^{12}\text{C}\) and \(^{14}\text{C}\)
C) \(\text{O}_{2}\) and \(\text{O}_{3}\)
D) \(\text{Na}^{+}\) and \(\text{Na}\)
Answer: B
Rationale: Isotopes are atoms of the same element with the
same number of protons (same atomic number) but
different numbers of neutrons (different mass numbers).
Carbon-12 and Carbon-14 both have 6 protons but differ in
neutrons.
2. What is the empirical formula of a compound that is
40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass?
A) \(\text{CHO}\)
B) \(\text{CH}_2\text{O}\)
,C) \(\text{C}_2\text{H}_4\text{O}_2\)
D) \(\text{CHO}_{2}\)
Answer: B
Rationale: Assuming a 100 g sample gives 40.0 g C, 6.7 g H,
and 53.3 g O. Converting to moles: \(40..01 = 3.33\text{
mol C}\), \(6..008 = 6.65\text{ mol H}\), and \(53..00
= 3.33\text{ mol O}\). Dividing by the smallest value (3.33)
yields a 1:2:1 ratio, resulting in \(\text{CH}_2\text{O}\).
3. How many protons, neutrons, and electrons are in a
\(^{37}\text{Cl}^-\) ion?
A) 17 protons, 20 neutrons, 17 electrons
B) 17 protons, 20 neutrons, 18 electrons
C) 18 protons, 19 neutrons, 17 electrons
D) 17 protons, 17 neutrons, 18 electrons
Answer: B
Rationale: Chlorine always has an atomic number of 17 (17
protons). The mass number is 37, so neutrons = \(37 - 17 =
20\). Because it has a -1 charge, it has gained one electron,
totaling 18 electrons.
4. Which element has the ground-state electron
configuration \([Ar] 4s^2 3d^{10} 4p^3\)?
A) \(\text{V}\)
B) \(\text{As}\)
C) \(\text{Se}\)
,D) \(\text{Sb}\)
Answer: B
Rationale: Counting the total electrons past Argon (18): \(18
+ 2 + 10 + 3 = 33\). The element with atomic number 33 on
the periodic table is Arsenic (\(\text{As}\)).
5. What is the molarity of a solution made by dissolving 4.0
grams of \(\text{NaOH}\) (molar mass = 40.0 g/mol) in
enough water to make 250 mL of solution?
A) 0.10 M
B) 0.40 M
C) 1.0 M
D) 0.25 M
Answer: B
Rationale: Moles of \(\text{NaOH} = 4.0\text{ g} / 40.0\text{
g/mol} = 0.10\text{ mol}\). Volume in liters = \(0.250\text{ L}\).
Molarity = \(0.10\text{ mol} / 0.250\text{ L} = 0.40\text{ M}\).
6. Identify the limiting reactant when 2.0 moles of
\(\text{H}_{2}\) react with 2.0 moles of \(\text{O}_{2}\) to form
water via the equation: \(2\text{H}_2(g) + \text{O}_2(g)
\rightarrow 2\text{H}_2\text{O}(g)\).
A) \(\text{H}_{2}\)
B) \(\text{O}_{2}\)
C) \(\text{H}_2\text{O}\)
D) Neither reactant is limiting
, Answer: A
Rationale: According to the stoichiometry, 2.0 moles of
\(\text{H}_{2}\) require \(1.0\text{ mole}\) of \(\text{O}_{2}\).
Since we have 2.0 moles of \(\text{O}_{2}\), \(\text{O}_{2}\) is
in excess and \(\text{H}_{2}\) will run out first, making it the
limiting reactant.
7. Which of the following molecules has a tetrahedral
molecular geometry?
A) \(\text{BF}_{3}\)
B) \(\text{NH}_{3}\)
C) \(\text{CH}_{4}\)
D) \(\text{SF}_{4}\)
Answer: C
Rationale: Methane (\(\text{CH}_{4}\)) has 4 bonding pairs
and 0 lone pairs around the central carbon atom, resulting
in a perfect tetrahedral geometry. \(\text{NH}_{3}\) is
trigonal pyramidal due to a lone pair.
8. What is the formal charge on the central nitrogen atom in
the nitrate ion (\(\text{NO}_{3}^{-}\))?
A) -1
B) 0
C) +1
D) +2
Answer: C
