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CHEM 1201 Exam 2 Form 1 | Questions and Answers | Update | 100% Correct.

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CHEM 1201 Exam 2 Form 1 | Questions and Answers | Update | 100% Correct.










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September 17, 2025
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Written in
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Exam 2 Form 1

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the
question.

1) How many Zn atoms are contained in 922 g of Zn?

A) 5.90 x 1025 Zn atoms
B) 7.09 x 1021 Zn atoms
C) 8.49 x 1024 Zn atoms
D) 4.18 x 1024 Zn atoms
E) 4.27 x 1022 Zn atoms




(
1molZn 6.022 % 1023 Zn atoms
922 gZn X ) =8.49 x 10** Zn at oms
g
65.39 ol 1molZn




2) What is the empirical formula of a substance that contains 5.28 g of C, 1.11 g of H, and 3.52 g of O?

A) CoH50
B) C3H404 Calculate moles of each element, then divide by the
C) C2H403
smallest number of moles: Oxygen (O)
D) C2H402
E) None of these _ 528g 0.440 mol C _
Moles of C = TYTE 0.440 mol 5220 molo
mol

111 1.10 mol H
Molesof H=——2 _~1.10mol ——— =
1.008 g/mol 0.220 mol O

3.52 0.220 mol O
Molesof 0 = ——9%— ~ 0.220 mol ———
16.00 g/mol 0.220 mol O

Empirical formula: C.HsO




3) What mass (in kg) does 4.77 moles of nickel have?

A) 0.280 kg
B) 0.352 kg Mass = moles X molar mass
C) 0.632 kg
D) 0.122 kg 58.693
g Ni 1 kg Ni
E) 0.820 kg m = 4.77 mot
mol Ni L= 1 mol Ni g Ni = 0.280 kg
" 1000

, 4) What is the coefficient for oxygen when the following equation is balanced using the lowest, whole-
number coefficients?

_
CgHgO2(l) + _ 0O2(g — ___ CO2(9) + ___ H20()

A)1 B)5 C)3 D)7 E) None of these
Balance reactants 2 Products
7
C3H602 + EOZ - 3602 + 3H20


Count oxygen in products: 9

Two oxygen comes from C3HgO2, which requires 9-2=7 more oxygen
70
2 2 =3.50
— - 2


Multiply all coefficients by 2 to remove fraction.
ZC3H602 + 702 - 6602 + 6H20




5) How many molecules of HCI are formed when 90.0 g of water reacts according to the following
balanced reaction? Assume excess ICI3.

2 1Cl3 + 3 H20 — ICI + HIO3 + 5 HCI

A) 3.00 x 1024 molecules HCI
B) 9.00 x 1024 molecules HCI
C) 5.00 x 1025 molecules HCI
D) 5.00 x 1024 molecules HCI
E) 6.00 x 1024 molecules HCI

1molH,0 5molHCl 6.022 x 1023 molecules
0gH
20.0gH;0* 18 01 g H,0" 3molH,0" 1mol HCL = 5.00 x 102*molecules HCI




6) How many moles of BCI3 are needed to produce 5.00 g of HCl(aq) in the following reaction?

BCI3 (g) + 3 H20 (I) — 3 HCI (aq) + B(OH)3 (aq)
A) 0.0457 mol
B) 0.411 mol
C) 21.9 mol
D) 0.137 mol Moles of BCl; =
E) None of these
1molHCl 1molBCl;
~ 0.0457 mol
5.00
g HCl* o 6 g HCl " 3 mol HCI

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