chapter 2 test version 3
1) elements in group 6a are known as the
a) alkali metals b) chalcogens c) alkaline earth metals d) halogens e) noble gases
2) elements in group 8a are known as the
a) halogens b) alkali metals c) alkaline earth metals d) chalcogens e) noble gases
3) an atom of the most common isotope of gold, 197au, has ______ protons,
______ neutrons, and ______ electrons.
a) 197, 79, 118 b) 118, 79, 39 c) 79, 197, 197 d) 79, 118, 118 e) 79, 118, 79
4) which combination of protons, neutrons, and electrons is correct for the isotope of
copper, $_{29}^{63}cu$?
a) 29 p+, 34 n°, 29 e- b) 29 p+, 29 n°, 63 e- c) 63 p+, 29 n°, 63 e- d) 34 p+, 29 n°, 34 e-
e) 34 p+, 34 n°, 29 e-
5) which isotope has 36 electrons in an atom?
a) $_{36}^{80}kr$ b) $_{35}^{80}br$ c) $_{34}^{78}se$ d) $_{17}^{34}cl$ e) $_{80}^{36}hg$
6) the formula weight of aluminum sulfate (al₂(so₄)₃) is ______ amu.
a) 342.15 b) 123.04 c) 59.04 d) 150.14 e) 273.06
7) the formula weight of silver chromate (ag₂cro₄) is ______ amu.
a) 159.87 b) 223.87 c) 331.73 d) 339.86 e) 175.87
8) the mass % of al in aluminum sulfate (al₂(so₄)₃) is ______.
a) 7.886 b) 15.77 c) 21.93 d) 45.70 e) 35.94
9) calculate the percentage by mass of nitrogen in ptcl₂(nh₃)₂.
a) 4.67 b) 9.34 c) 9.90 d) 4.95 e) 12.67
10) what is the empirical formula of a compound that contains 29% na, 41% s, and 30%
o by mass?
a) na₂s₂o₃ b) naso₂ c) naso d) naso₃ e) na₂s₂o₆
11) what is the empirical formula of a compound that contains 49.4% k, 20.3% s, and
30.3% o by mass?
a) kso₂ b) kso₃ c) k₂so₄ d) k₂o₃ e) kso₄
12) combustion of a 1.031 - g sample of a compound containing only carbon, hydrogen, and
oxygen produced 2.265 g of co₂ and 1.236 g of h₂o. what is the empirical formula of the
compound?
a) c₃h₆o b) c₃h₅o c) c₆h₁₆o₂ d) c₃h₉o₃ e) c₃h₆o₃
13) a compound that is composed of only carbon and hydrogen contains 80.0% c and
20.0% h by mass. what is the empirical formula of the compound?
a) c₂₀h₄₀ b) c₇h₂₀ c) ch₃ d) c₂h₆ e) ch₄

Question

Accepted Answer

1. **Answer - Question 1**:
   - # Brief Explanations: Elements in Group 6A are called chalcogens. Halogens are in Group 7A, alkali - metals in Group 1A, alkaline earth metals in Group 2A, and noble gases in Group 8A.
   - # Answer: B. chalcogens
2. **Answer - Question 2**:
   - # Brief Explanations: Elements in Group 8A are known as noble gases. Halogens are in Group 7A, alkali - metals in Group 1A, alkaline earth metals in Group 2A, and chalcogens in Group 6A.
   - # Answer: E. noble gases
3. **Answer - Question 3**:
   - # Brief Explanations: The atomic number of gold (Au) is 79, so the number of protons is 79. For a neutral atom, the number of electrons equals the number of protons, so there are 79 electrons. The mass number of $^{197}Au$ is 197, and the number of neutrons = mass number - atomic number = 197 - 79=118.
   - # Answer: E. 79, 118, 79
4. **Answer - Question 4**:
   - # Brief Explanations: For $_{29}^{63}Cu$, the atomic number is 29, so the number of protons ($p^+$) is 29. In a neutral atom, the number of electrons ($e^-$) is also 29. The mass number is 63, and the number of neutrons ($n^0$) = mass number - atomic number = 63 - 29 = 34.
   - # Answer: A. $29\ p^+,34\ n^0,29\ e^-$
5. **Answer - Question 5**:
   - # Brief Explanations: In a neutral atom, the number of electrons equals the atomic number. For $^{80}_{36}Kr$, the atomic number is 36, so it has 36 electrons.
   - # Answer: A. $^{80}_{36}Kr$
6. **Answer - Question 6**:
   - # Explanation:
     - ## Step1: Calculate the atomic weights of elements in $Al_2(SO_4)_3$. Atomic weight of $Al = 26.98\ amu$, $S=32.07\ amu$, $O = 16.00\ amu$.
     - The formula weight $=2\times26.98+3\times(32.07 + 4\times16.00)$
     - ## Step2: First, calculate the value inside the parentheses: $32.07+4\times16.00=32.07 + 64.00=96.07$.
     - Then, $3\times96.07 = 288.21$ and $2\times26.98=53.96$.
     - The formula weight $=53.96+288.21 = 342.17\approx342.15\ amu$.
   - # Answer: A. 342.15
7. **Answer - Question 7**:
   - # Explanation:
     - ## Step1: Atomic weight of $Ag = 107.87\ amu$, $Cr = 52.00\ amu$, $O=16.00\ amu$.
     - The formula weight of $Ag_2CrO_4=2\times107.87+52.00 + 4\times16.00$.
     - ## Step2: $2\times107.87 = 215.74$, $4\times16.00 = 64.00$.
     - Then $215.74+52.00+64.00=331.74\approx331.73\ amu$.
   - # Answer: C. 331.73
8. **Answer - Question 8**:
   - # Explanation:
     - ## Step1: Formula weight of $Al_2(SO_4)_3\approx342.15\ amu$ (from question 6), and the mass of $Al$ in $Al_2(SO_4)_3$ is $2\times26.98 = 53.96\ amu$.
     - ## Step2: Mass $\%$ of $Al=\frac{53.96}{342.15}\times100\%\approx15.77\%$.
   - # Answer: B. 15.77
9. **Answer - Question 9**:
   - # Explanation:
     - ## Step1: Calculate the formula weight of $PtCl_2(NH_3)_2$. Atomic weights: $Pt = 195.08\ amu$, $Cl=35.45\ amu$, $N = 14.01\ amu$, $H = 1.01\ amu$.
     - Formula weight $=195.08+2\times35.45+2\times(14.01 + 3\times1.01)$
     - $=195.08 + 70.90+2\times(14.01+3.03)$
     - $=195.08 + 70.90+2\times17.04$
     - $=195.08+70.90 + 34.08=299.96\ amu$.
     - The mass of $N$ in $PtCl_2(NH_3)_2$ is $2\times14.01 = 28.02\ amu$.
     - ## Step2: Mass $\%$ of $N=\frac{28.02}{299.96}\times100\%\approx9.34\%$.
   - # Answer: B. 9.34
10. **Answer - Question 10**:
   - # Explanation:
     - ## Step1: Assume 100 g of the compound. So, $m_{Na}=29\ g$, $m_S = 41\ g$, $m_O=30\ g$.
     - Calculate the number of moles: $n_{Na}=\frac{29\ g}{22.99\ g/mol}\approx1.26\ mol$, $n_S=\frac{41\ g}{32.07\ g/mol}\approx1.28\ mol$, $n_O=\frac{30\ g}{16.00\ g/mol}=1.875\ mol$.
     - ## Step2: Divide by the smallest number of moles ($n_{Na}\approx1.26\ mol$).
     - $Na:S:O=\frac{1.26}{1.26}:\frac{1.28}{1.26}:\frac{1.875}{1.26}\approx1:1:1.5$. Multiply by 2 to get whole - numbers: $Na_2S_2O_3$.
   - # Answer: A. $Na_2S_2O_3$
11. **Answer - Question 11**:
   - # Explanation:
     - ## Step1: Assume 100 g of the compound. So, $m_K = 49.4\ g$, $m_S=20.3\ g$, $m_O = 30.3\ g$.
     - Calculate the number of moles: $n_K=\frac{49.4\ g}{39.10\ g/mol}\approx1.26\ mol$, $n_S=\frac{20.3\ g}{32.07\ g/mol}\approx0.63\ mol$, $n_O=\frac{30.3\ g}{16.00\ g/mol}\approx1.89\ mol$.
     - ## Step2: Divide by the smallest number of moles ($n_S\approx0.63\ mol$).
     - $K:S:O=\frac{1.26}{0.63}:\frac{0.63}{0.63}:\frac{1.89}{0.63}=2:1:3$, so the empirical formula is $K_2SO_4$.
   - # Answer: C. $K_2SO_4$
12. **Answer - Question 12**:
   - # Explanation:
     - ## Step1: Calculate the moles of $C$ from $CO_2$ and $H$ from $H_2O$.
     - Moles of $C$ in $2.265\ g\ of\ CO_2$: $n_C=\frac{2.265\ g}{44.01\ g/mol}=0.0515\ mol$.
     - Moles of $H$ in $1.236\ g\ of\ H_2O$: $n_H = 2\times\frac{1.236\ g}{18.02\ g/mol}=0.137\ mol$.
     - Mass of $C=0.0515\ mol\times12.01\ g/mol = 0.618\ g$, mass of $H=0.137\ mol\times1.01\ g/mol=0.138\ g$.
     - Mass of $O$ in the compound $=1.031-(0.618 + 0.138)=0.275\ g$.
     - Moles of $O=\frac{0.275\ g}{16.00\ g/mol}=0.0172\ mol$.
     - ## Step2: Divide by the smallest number of moles ($n_O\approx0.0172\ mol$).
     - $C:H:O=\frac{0.0515}{0.0172}:\frac{0.137}{0.0172}:\frac{0.0172}{0.0172}\approx3:8:1$, so the empirical formula is $C_3H_8O$.
   - # Answer: B. $C_3H_8O$
13. **Answer - Question 13**:
   - # Explanation:
     - ## Step1: Assume 100 g of the compound. So, $m_C = 80.0\ g$, $m_H=20.0\ g$.
     - Moles of $C=\frac{80.0\ g}{12.01\ g/mol}\approx6.66\ mol$, moles of $H=\frac{20.0\ g}{1.01\ g/mol}\approx19.8\ mol$.
     - ## Step2: Divide by the smallest number of moles ($n_C\approx6.66\ mol$).
     - $C:H=\frac{6.66}{6.66}:\frac{19.8}{6.66}\approx1:3$, so the empirical formula is $CH_3$.
   - # Answer: C. $CH_3$

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