CHEM101- Final Exam- Version A- Deatiled Answers

CHEM MISC


General Chemistry 1(CHEM101) Final Exam (A) Fall2021
Duration: 2h 00 min
Student Name:______________________________ Student ID: ___________________
Professor’s Name:___________________________ Section: L_____
Instructions:
1. The exam contains 35 Multiple Choice Questions (Each question is 1 point, 35
points total)
2. Answers should be entered in the “Answers Sheet” (page#5)
3. A simple periodic Table is provided with the exam copy.
Constants:
- Avogadro’ s number: NA = 6.021023 atoms/mol
- Gas constant: R = 0.0821 L.atm/mol.K
- 1 atm = 760 mmHg = 760 torr = 101325 Pa
- T(K) = T(°C) +273
- Planck’s constant: h = 6.6310-34 J.s
- Light speed (celerity): c = 3.0108 m/s
−𝟐.𝟏𝟖×𝟏𝟎−𝟏𝟖
- Bohr’ s model: 𝑬𝒏 = 𝑱
𝒏𝟐
A1. How many significant figures in the following number: 3.050  10-3 m?
A. 2 B. 3 C. 4
D. 5 E. None of the above
The number of SF in 3.050  10-3, is the same than in 3.050. For numbers greater than 1, all the
zeros to the right of the decimal point are significant. So, 3.050  10-3 contains 4 SF.
A2. How many cm is in 123 nm?
A. 1.2310-1 cm B. 1.2310-2 cm C. 1.2310-3 cm
-5
D. 1.2310-4 cm E. 1.2310 cm
Knowing that: 1 nm = 10-9 m and 1 cm = 10-2 m, 123 nm is converted to cm using a two-
step dimensional analysis:
10−9 𝑚 1 𝑐𝑚
123 𝑛𝑚 × × −2 = 1.23 × 10(2−9+2) 𝑐𝑚 = 1.23 × 10−5 𝑐𝑚
1𝑛𝑚 10 𝑚
A3. Carry out the following calculation to the correct number of significant figures:
3.463 × 3.17 =
A. 10.97771 B. 10.9777 C. 10.978
D. 10.98 E. 11.0
For a multiplication, the result should be rounded to a number having the smallest number of SF
of the members of the calculation (3 SF in 3.17): 3.463 × 3.17 = 10.97771. The result should be
converted to 3 SF: 11.0
A4. The name of Na3N is:
A. Sodium nitrate B. Sodium nitride C. Sodium nitrite
D. Trisodium nitrate E. Trisodium nitrite
Na3N is an ionic compound containing sodium as a alkaline metal cation (Na+) and monoatomic
anion from nitrogen (N3-). An ionic compound is named first by its cation and then by its anion.
Name first the metal and then the non-metal adding the “ide” to the root of the element nitrogen:
Sodium nitride.



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, A.5 The formula of tetraphosphorus decoxide is:
A. P2O5 B. P2O3 C. P4O10
D. P3O7 E. P8O20
Tetraphosphorus decoxide is a molecular compound containing four phosphorus atoms
(the prefix “tetra” designates 4) and ten oxygen atoms (the prefix “deca” designates 10)
in its molecular formula. Thus, the molecular formula is P4O10.
A6. The formula of sodium hydrogen tetraborate pentahydrate is (H2B4O7 is tetraboric acid):
A. Na2B4O5. 4H2O B. NaHB4O7. 5H2O C. Na3B4O6. 7H2O
D. Na3HB4O7. 5H2O E. Na2HB4O7. 5H2O
H2B4O7 is tetraboric acid, then B4O72- is tetraborate and HB4O7- is hydrogen tetraborate. sodium
hydrogen tetraborate is NaHB4O7 and sodium hydrogen tetraborate pentahydrate (pentahydrate
indicates 5 H2O attached to the ionic compound) is: NaHB4O7. 5H2O.
A7. Write a balanced equation to show the reaction of gaseous ethane (C2H6) with gaseous
dioxygen (O2) to form carbon monoxide gas and water vapor.
A. 2 C2H6(g) + 7 O2(g) → 4 CO2 (g) + 6 H2O(g)
B. C2H6(g) + 5 O(g) → 2 CO(g) + 3 H2O(g)
C. 2 C2H6(g) + 5 O2(g) → 4 CO(g) + 6 H2O(g)
D. C2H6(g) + 7 O(g) → 2 CO2(g) + 3 H2O(g)
E. 2 CH3 (g) + 5 O(g) → 2 CO(g) + 3 H2O(g)
A8. Calcium phosphate reacts with sulfuric acid to form calcium sulfate and phosphoric acid
according to the balanced equation: Ca3(PO4)2(s) + 3 H2SO4(l)  3 CaSO4(s) + 2 H3PO4(l).
How many grams of H3PO4 can be formed from the reaction of 6.20 g Ca3(PO4)2(s) with excess
H2SO4. Molar masses: Ca3(PO4)2: 310 g/mol, H3PO4: 98 g/mol.
A. 1.96 g H3PO4 B. 3.92 g H3PO4 C. 7.84 g H3PO4
D. 5.88 g H3PO4 E. 6.20 g H3PO4
1 𝑚𝑜𝑙 𝐶𝑎3 (𝑃𝑂4 )2 2 𝑚𝑜𝑙 𝐻3 𝑃𝑂4 98 𝑔 𝐻3 𝑃𝑂4
6.20 𝑔 𝐶𝑎3 (𝑃𝑂4 )2 × × × = 3.92 𝑔𝐻3 𝑃𝑂4
310 𝑔 𝐶𝑎3 (𝑃𝑂4 )2 1 𝑚𝑜𝑙 𝐶𝑎3 (𝑃𝑂4 )2 1 𝑚𝑜𝑙 𝐻3 𝑃𝑂4
A9. Select the hydrogen displacement reaction from the following:
A. MgCl2(s) + Ca(s)  CaCl2(s) + Mg(l)
B. 2Ca(l) + O2(g)  2 CaO(s)
C. CaCl2(aq) + Br2(l)  CaBr2(aq) + Cl2(g)
D. 3Ca(s) + 2H3PO4 (aq)  Ca3(PO4)2(s) + 3H2(g)
E. CaF2(s)  Ca(s) + F2(g)
A10. 0.3455 g of A2CO3 (A is an alkaline metal) was reacted with 50.0 mL 0.200 M hydrochloric
acid (HCl) to form ACl(aq), H2O(l) and CO2(g): M2CO3(aq)+2HCl(aq)  2ACl(aq)+CO2(g)+H2O(l).
The excess of HCl requires 50.0 mL of 0.100 M NaOH aqueous solution. After determining the
molar mass of A2CO3, identify A.
A. Li B. Cs C. K
D. Na E. Rb
The balanced reaction between A2CO3 and HCl is given by the equation:
𝐴2 𝐶𝑂3 (𝑎𝑞) + 2𝐻𝐶𝑙 → 2𝑀𝐶𝑙(𝑎𝑞) + 𝐶𝑂2 (𝑔) + 𝐻2 𝑂(𝑙)
𝑟𝑒𝑎𝑐𝑡𝑒𝑑
𝑛𝐻𝐶𝑙
Based on the stoichiometry of the reaction: 𝑛𝐴2 𝐶𝑂3 = 2
𝑟𝑒𝑎𝑐𝑡𝑒𝑑 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑒𝑥𝑐𝑒𝑠𝑠
𝑛𝐻𝐶𝑙 = 𝑛𝐻𝐶𝑙 − 𝑛𝐻𝐶𝑙
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
1𝐿
𝑛𝐻𝐶𝑙 = 𝑀𝐻𝐶𝑙 × 𝑉𝐻𝐶𝑙 = 0.200 𝑀 × 50.0 𝑚𝐿 × = 0.010 𝑚𝑜𝑙
1000 𝑚𝐿


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