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chem 120 final questions and answers to week 8

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Week 8 : Final Exam Questions - Final Exam CHEM120 Final-Exam- 1. Question: (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 0.25L of a 0.2M H3PO4 solution. Show your work. (b, 5 pts) What volume, in Liters, of a 0.2 M H3PO4 solution can be prepared by diluting 50 mL of a 5M H3PO4 solution? Show your work. (Pts. : 10) A. Molarity = moles of solute/liters of solution moles of solute = 0.2 M*0.25 L = 0.05 mol H3PO4 Using the molar mass given, convert this amount to grams. mass = 0.05 mol * (97.994 g/mol) = 4.89 grams H3PO4 B. C1V1 = C2V2. C1 = 5 M, V1=0.05L, C2 = 0.2M; V2 = [(5M)(0.05L)]/(0.2M) = 1.25L 2. Question: (TCO 7) (a, 5 pts) What is the mass/volume percent of a solution prepared by dissolving 43 g of NaOH in enough water to make a final volume of 120 mL? Show your work. (b, 5 pts) How many mL of a 10% solution can be made from the solution in part a? Show your work. (Pts. : 10) 3. Question: (TCO 12) Polyethylene is a polymer found in many applications, including packaging for fruit and vegetables. Discuss the structural differences between (1) polyethylene, (2) polypropylene, and (3) polystyrene and how the structure impacts their commercial uses. (Pts. : 15) instruments and appliances, and it is w idely used for home insulation. 4. Question: (TCO 11) Tungsten (W), with a mass number of 180 and an atomic number of 74, decays by emission of an alpha particle. Identify the product of the nuclear reaction by providing its atomic symbol (5 pts), mass number (5 pts), and atomic number (5 pts). (Pts. : 15) 6. Question: (TCO 13) What is the mRNA sequence for the following segment of DNA: --TAACGAATAGCCTGT-- (10 pts)? Based upon the mRNA sequence, what is the peptide sequence (10 pts)? (Pts. : 20) DNA RNA A = U T = A C = G G = C TAACGAATAGCCTGT w ill become AUUGCUUAUCGGACA Peptide sequence is AUU GCU UAU CGG ACA w hich is Ile-Ala-Tyr-Arg-Thr

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Week 8 : Final Exam
Questions - Final Exam
CHEM120 Final-Exam-



1. Question: (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 0.25L of a 0.2M H3PO4
solution. Show your work.
(b, 5 pts) What volume, in Liters, of a 0.2 M H3PO4 solution can be prepared by diluting 50 mL
of a 5M H3PO4 solution? Show your work. (Pts. : 10)


A. Molarity = moles of solute/liters of
solution moles of solute = 0.2 M*0.25 L =
0.05 mol H3PO4

Using the molar mass given, convert this amount to
grams. mass = 0.05 mol * (97.994 g/mol) = 4.89
grams H3PO4




B. C1V1 = C2V2. C1 = 5 M, V1=0.05L, C2 = 0.2M; V2 = [(5M)
(0.05L)]/(0.2M) = 1.25L




2. Question: (TCO 7) (a, 5 pts) What is the mass/volume percent of a solution prepared by
dissolving 43 g of NaOH in enough water to make a final volume of 120 mL? Show your
work.
(b, 5 pts) How many mL of a 10% solution can be made from the solution in part a? Show your
work. 2. First convert the given mass of NaOH to volume (in mL) using the
density of NaOH w hich is 2.13 g/mL.
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:

V_total = 201.88 mL
So, about 202 mL of a 10% solution can be made from the solution in part
a.




(Pts. : 10)

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