CHEM 120 Week 8 Final Exam
Course Number:CHEM120
Course Title:Introduction to General,Organic & Biological Chemistry with Lab
Correct Question & Answers
Secure high grade
Best feedback (rated) document by students
,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
0 1644446450 Short 1
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
0 1644446452 Short 6
3. (TCO 3) What is the name of the following compound: Zn3P2? (Points : 5)
It's Zinc Phosphide
0 1644446453 Short 7
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)
Using Charles’ Law, (V1/T1) = (V2/T2).
First, convert temperature to KELVIN (T1 = t1 +273)
Thus, T1 = 95 + 273 = 368.
We have V1 (165 mL) & T2 = (25 + 273) = 298.
V2 = (V1*T2)/T1 = (165 mL*298)/368 = 133.6 mL.
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7. (TCO 6) A sample of helium gas occupies 1021 mL at 719 mmHg. For a gas sample at constant
temperature, determine the volume of helium at 745 mmHg. Show your work. (Points : 5)
1021mL * 719 mm/745 mmHg = 985.36mL =985mL
Using Boyle’s law, P1V1 = P2V2. We have V1 (1021 mL), P1 (719 mmHg) and P2 (745
mmHg).
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8. (TCO 12) If one strand of a DNA double helix has the sequence T T A G C G A C G C, what is the
sequence of the other DNA strand? (Points : 10)
AATCGCTGCG
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1. (TCO 7) (a, 5 pts) Given that the molar mass of H3PO4 is 97.994 grams, determine the number of
grams of H3PO4 needed to prepare 0.75L of a 0.25M H3PO4 solution. Show your work.
(b, 5 pts) What volume, in Liters, of a 0.25 M H3PO4 solution can be prepared by diluting 50 mL of a 2.5M
H3PO4 solution? Show your work. (Points : 10)
Using the molar mass given, convert this amount to grams.
mass = 0.1875 mol * (97.994 g/mol) = 18.37 grams H3PO4
b. M1*V1 = M2*V2
w here: M1 = 0.25; V1 = ??; M2 = 2.5; V2 = 50 mL = 0.050 L
Solvig for V1:
0.25 * V1 = 2.5 * 0.050
V1 = 0.50 L
2. First convert the given mass of NaOH to volume (in mL) using the density of NaOH
Volume = 43 g * (1 mL/2.13 g) = 20.19 mL
Volume % = (volume of solute / volume of solution) * 100%
Volume % = (20.19 mL/120 mL) * 100% = 16.8 %
b. Volume % = volume of NaOH/ total volume
0.10 = 20.19 mL/total volume
Solving for total volume yields:
0 1644446463 Essay 3
2. (TCO 7) (a, 5 pts) What is the volume percent of a solution prepared by dissolving 21 g of NaOH in
Course Number:CHEM120
Course Title:Introduction to General,Organic & Biological Chemistry with Lab
Correct Question & Answers
Secure high grade
Best feedback (rated) document by students
,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
0 1644446450 Short 1
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
0 1644446452 Short 6
3. (TCO 3) What is the name of the following compound: Zn3P2? (Points : 5)
It's Zinc Phosphide
0 1644446453 Short 7
4. (TCO 3) What is the name of the following compound: AgNO3? (Points : 5)
Silver nitrate
0 1644446454 Short 11
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)
Using Charles’ Law, (V1/T1) = (V2/T2).
First, convert temperature to KELVIN (T1 = t1 +273)
Thus, T1 = 95 + 273 = 368.
We have V1 (165 mL) & T2 = (25 + 273) = 298.
V2 = (V1*T2)/T1 = (165 mL*298)/368 = 133.6 mL.
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7. (TCO 6) A sample of helium gas occupies 1021 mL at 719 mmHg. For a gas sample at constant
temperature, determine the volume of helium at 745 mmHg. Show your work. (Points : 5)
1021mL * 719 mm/745 mmHg = 985.36mL =985mL
Using Boyle’s law, P1V1 = P2V2. We have V1 (1021 mL), P1 (719 mmHg) and P2 (745
mmHg).
0 1644446459 Short 19
8. (TCO 12) If one strand of a DNA double helix has the sequence T T A G C G A C G C, what is the
sequence of the other DNA strand? (Points : 10)
AATCGCTGCG
0 1644446456 Short 14
1. (TCO 7) (a, 5 pts) Given that the molar mass of H3PO4 is 97.994 grams, determine the number of
grams of H3PO4 needed to prepare 0.75L of a 0.25M H3PO4 solution. Show your work.
(b, 5 pts) What volume, in Liters, of a 0.25 M H3PO4 solution can be prepared by diluting 50 mL of a 2.5M
H3PO4 solution? Show your work. (Points : 10)
Using the molar mass given, convert this amount to grams.
mass = 0.1875 mol * (97.994 g/mol) = 18.37 grams H3PO4
b. M1*V1 = M2*V2
w here: M1 = 0.25; V1 = ??; M2 = 2.5; V2 = 50 mL = 0.050 L
Solvig for V1:
0.25 * V1 = 2.5 * 0.050
V1 = 0.50 L
2. First convert the given mass of NaOH to volume (in mL) using the density of NaOH
Volume = 43 g * (1 mL/2.13 g) = 20.19 mL
Volume % = (volume of solute / volume of solution) * 100%
Volume % = (20.19 mL/120 mL) * 100% = 16.8 %
b. Volume % = volume of NaOH/ total volume
0.10 = 20.19 mL/total volume
Solving for total volume yields:
0 1644446463 Essay 3
2. (TCO 7) (a, 5 pts) What is the volume percent of a solution prepared by dissolving 21 g of NaOH in
