CHEM 120, Final Exam Chamberlain college of Nursing,CHEM 120: Introduction to General, Organic & Biological Chemistry Set-1

CHEM: 120 Final Exam
1. (TCO 8) 35.0 mL of 0.25 M NaOH is neutralized by 23.6 mL of an HCl solution. The molarity of the HCl
solution is (show your work): (Points : 5)

molarity = moles solute / liters solution
0.25 M = moles NaOH / 0.035 L
moles NaOH = 0.00875 moles NaOH




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2. (TCO 1) How many mL are in 3.5 pints? Show your work. (Points : 5)

3.5 pints is equivalent to 1656.116

1 pint = 473.176 ml
3.5 pints* 473.176mL = 1656.116mL


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3. (TCO 3) What is the name of the following compound: Zn3P2? (Points : 5)

It's Zinc Phosphide




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4. (TCO 3) What is the name of the following compound: AgNO3? (Points : 5)

Silver nitrate




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5. (TCO 6) Calculate the pressure, in atmospheres, of 2.78 mol CO(g) in a 4.25 L tank at 51 degrees C.
(Points : 5)

Given that n = 2.78 mol; V = 4.25 L; and temperature

[2.78 mol* 0.0821 L -atm/ mol-K*304 K)/4.25 L = 16.3




6. (TCO 6) A gas at a temperature of 95 degrees C occupies a volume of 165 mL. Assuming constant
pressure, determine the volume at 25 degrees C. Show your work. (Points : 5)