Download CHEM 120 FINAL EXAM||2025-2026|| (VERSION-2) QUESTIONS WITH CORRECT ANSWERS and more Exams Chemistry in PDF only on Docsity!
0 1644446450 Short 1 molarity = moles solute / liters solution 0.25 M = moles NaOH / 0.035 L moles NaOH = 0.00875 moles NaOH 3.5 pints is equivalent to 1656. 1 pint = 473.176 ml 3.5 pints* 473.176mL = 1656.116mL 0 1644446452 Short^6
CHEM 120 FINAL EXAM||2025-2026|| (VERSION-2)
QUESTIONS WITH CORRECT ANSWERS, WELL
ELABORATE WORKINGS AND CALCULATIONS. A+
GRADE
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 = (V1T2)/T1 = (165 mL298)/368 = 133.6 mL.
0 1644446457 Short 16
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 (
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)
A A T C G C T G C G
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)
2. (TCO 1) How many mL are in 3.5 pints? Show your work. (Points : 5)
3. (TCO^ 3)^ What^ is^ the^ name^ of^ the^ following^ compound:^ Zn 3 P 2?^ (Points^ : 5)
0 1644446464 Essay 5 Part 1: based on the density of NAOH =2.1g/mL then calculate the volume of 21g NaOHx(1mL/2.1g)=10 mL so the percent volume is defined as % volume = (Volume of Solute/volume of solution) x 100 so w e have % volume = (10mL/120mL)x100=8.3% Part 2: you cannot prepare 10 % (volume) from the above solution (8.3%) because the f inal solution is more concentrated than the initial 0 1644446467 Essay 11 mass of 180 becomes 176 atomic number of 74 becomes 72 name: Hafnium Symbol: Hf 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
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.
(Points : 10)
0 1644446466 Essay 7
4. (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). (Points : 15)
0 1644446470 Essay 16 DNA RNA A = U T = A C = G G = C AAACGTGTGCTAACA w ill become UUUGCACACGAUUGU Peptide sequence is UUU GCA CAC GAU UGU w hich is Phe-Ala-His-Asp-Cys 0 1644446469 Essay 14
6. (TCO 13) What is the mRNA sequence for the following segment of DNA:
--AAACGTGTGCTAACA-- (10 pts)? Based upon the mRNA sequence, what is the peptide sequence ( pts)?
(Points : 20)
7. (TCO 5) Given the following unbalanced chemical equation: Al + Cl 2 - > AlCl 3
(a, 5 pts) Balance the equation. (b, 5 pts) How many moles of AlCl 3 are produced from 1.75 mole of Cl 2? Show your work. (c, 5 pts) What is the molar mass of AlCl 3? Show your work.
(d, 5 pts) Calculate the number of grams of AlCl 3 produced from 1.75 mol Cl 2. Show your work. (Points :
transfer RNA. ribosomal RNA. messenger RNA. a new cell. 0 1644446422 MultipleChoice 5
2. (TCO 12) The portion of an enzyme where the substrate “fits” during the reaction is called the
(Points : 5)
active site. action site. reaction site. substrate site. inhibitor site. 0 1644446423 MultipleChoice 8
3. (TCO 12) Which of the following is not an unsaturated fatty acid? (Points : 5)
CH 3 CH 2 CH 2 CH=CHCOOH
CH 3 CH=CHCH=CHCOOH
CH 3 CH 2 CH 2 CH 2 CH 2 COOH
CH 3 CH 2 CH 2 CH 2 CH 2 CH=CHCOOH
CH 3 CH=CHCH 2 CH=CHCOOH
0 1644446425 MultipleChoice 10
4. (TCO 12) The helical structure of certain proteins, such as wool, is a part of the protein’s (Points : 5)
primary structure. secondary structure.
3 3 Lactose 0 1644446433 MultipleChoice 23
8. (TCO 4) The best description of the shape of an ammonia molecule is. (Points : 5)
linear pyramidal bent tetrahedral None of the above 0 1644446434 MultipleChoice 27
9. (TCO 4) Which of the following substances does not have polar covalent bonds? (Points : 5)
CO 2
Br 2 CH 4 NF 3 0 1644446437 MultipleChoice 28
10. (TCO 8) According to the Bronsted-Lowry definition, which chemical in the following reaction is the
acid?
H 2 CO 3 + H 2 O - > H 3 O+^ + HCO -^ (Points : 5)
H 2 O
HCO -
H 2 CO 3
H 3 O+
None of the above 0 1644446439 MultipleChoice 31
5 : Pineapple flavor
Answer
: Dichlorodiphenyltrichloroethane
4 1
: Formaldehyde
(^1 )
: Acetone
2 3
: Caffeine
: Ethyl butyrate
0 1644446445 Matching 5
14. (TCO 9) Match the organic compound with its name.
(Points : 15)
Potential Matches:
1 : Ethanoic acid
2 : 1 - propanol
3 : Ethylamine
4 : Heptanal
5 : Pentyl acetate
Answer
: CH 3 COOH
: CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CHO
: CH 3 CH 2 CH 2 OH
: CH 3 COOCH 2 CH 2 CH 2 CH 2 CH 3
: CH 3 CH 2 NH 2
Week 8 : Final Exam - Final Exam
CHEM120 Final Exam Page 3
1. (TCO 7) (a, 5 pts) Given that the molar mass of H 3 PO 4 is 97.994 grams, determine the number of 3 4 5 5 1 1 4 2 2 3 5 4 3 5
grams of H 3 PO 4 needed to prepare 0.25L of a 0.2M H 3 PO 4 solution. Show your work. (b, 5 pts) What volume, in Liters, of a 0.2 M H 3 PO 4 solution can be prepared by diluting 50 mL of a 5M H 3 PO 4 solution? Show your work. (Points : 10) A. Molarity = moles of solute/liters of solution moles of solute = 0.2 M*0.25 L = 0.05 mol H3PO Using the molar mass given, convert this amount to grams. mass = 0.05 mol * (97.994 g/mol) = 4.89 grams H3PO B. C1V1 = C2V2. C1 = 5 M, V1=0.05L, C2 = 0.2M; V2 = [(5M)(0.05L)]/(0.2M) = 1.25L
2. (^) (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. (Points : 10) 3. (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. (Points : 15) 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.
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
7. (TCO^ 5)^ Given^ the^ following^ unbalanced^ chemical^ equation:^ Al^ +^ Cl 2 - >^ AlCl 3 (a, 5 pts) Balance the equation. (b, 5 pts) How many moles of AlCl 3 are produced from 1.75 mole of Cl 2? Show your work. (c, 5 pts) What is the molar mass of AlCl 3? Show your work. (d, 5 pts) Calculate the number of grams of AlCl 3 produced from 1.75 mol Cl 2. Show your work. (Points :
- A. Balance: 2Al + 3 Cl2 > 2 AlCl B. Moles: 1.75 mol Cl2 * (2 mol AlCl3 / 3 mol Cl2) = 1.17 mol AlCl C. Molar Mass: Al = 26.981 g/mol Cl = 35.453 g/mol (1 x Al) + (3 x Cl) = (1 x 26.981g/mol) + (3 x 35.453g.mol) 26.981g/mol + 106.359 g/mol = 133.340g/mol = 133.34g/mol D. Grams: 1.75 mol cl2*2/133.34=0.026g 8. (TCO 13) What is the mRNA sequence for the following segment of DNA: --CTCGTGGTTTCATCC-- (10 pts)? Based upon the mRNA sequence, what is the peptide sequence ( pts)?
molarity = moles solute / liters solution 0.75 M = moles NaOH / 0.045 L moles NaOH = 0.03375 moles NaOH (Points : 20) DNA RNA A = U T = A C = G G = C CTCGTGGTTTCATCC w ill become GAGCACCAAAGUAGG Peptide sequence is GAG CAC CAA AGU AGG w hich is Glu-His-Gln-Ser-Arg
Week 8 : Final Exam - Final Exam
CHEM120 Final Exam Page 2
1. (TCO 8) 45.0 mL of 0.75 M NaOH is neutralized by 53.6 mL of an HCl solution. The molarity of the HCl solution is (show your work): (Points : 5) 2. (TCO 1) How many meters are in 175 inches? Show your work. (Points : 5)
TGCAGTACCG
8. (TCO 12) If one strand of a DNA double helix has the sequence ACGTCATGGC, what is the sequence of the other DNA strand? (Points : 10) 1. (TCO 12) The DNA double helix is. (Points : 5) made up of two polynucleotide strands composed of adenine and thymine a puzzle to geneticists composed of adenine and guanine composed of two chromosomes 2. (TCO 12) Which of the following best describes the action of enzymes in living systems? A specific enzyme generally catalyzes (Points : 5) one specific reaction. a group of similar reactions. many different reactions. either one specific reaction or a group of similar reactions. random reactions. 3. (TCO 12) Which of the following is not an unsaturated fatty acid? (Points : 5) CH 3 CH 2 CH 2 CH=CHCOOH CH 3 CH=CHCH=CHCOOH CH 3 CH 2 CH 2 CH 2 CH 2 COOH CH 3 CH 2 CH 2 CH 2 CH 2 CH=CHCOOH CH 3 CH=CHCH 2 CH=CHCOOH Using Boyle’s law , P1V1 = P2V2. We have V (1045 mL), P1 (721 mmHg) and P2 (745 mmHg). We w ant to determine V2, so V2 = (P1 x V1)/P2 = (1045 mL * 721 mmHg)/745 mmHg) = 1011 mL.
4. (TCO 12) The primary structure of a protein is determined by (Points : 5) the amino-acid composition. the order of amino acids in the protein. the hydrogen bonding that gives the protein a three-dimensional shape. the intertwining of protein molecules to form a “functional” protein. the hydrogen bonds between sulfur atoms on different amino acids. 5. (TCO 12) Which of the following is a disaccharide? (Points : 5) Cellulose Starch Ribose Galactose Lactose 6. (TCO 12) Cell nutrients and waste must pass through the (Points : 5) nuclear membrane. ribosomes. mitochondria chloroplasts. cell membrane. 7. (TCO 12) Which of the following is a monosaccharide? (Points : 5) Mannose Starch Cellulose Glucose Lactose 8. (TCO 4) The best description of the shape of an ammonia molecule is. (Points : 5) linear pyramidal
