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CHEM 120 FINAL EXAM||2025-2026|| (VERSION-2) QUESTIONS WITH CORRECT ANSWERS, WELL ELABORATE WORKINGS AND CALCULATIONS. A+ GRADE
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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
0 1644446457 Short 16
0 1644446459 Short 19
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
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.
0 1644446466 Essay 7
of an alpha particle. Identify the product of the nuclear reaction by providing its atomic symbol (5 pts),
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
--AAACGTGTGCTAACA-- (10 pts)? Based upon the mRNA sequence, what is the peptide sequence ( pts)?
(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.
transfer RNA. ribosomal RNA. messenger RNA. a new cell. 0 1644446422 MultipleChoice 5
active site. action site. reaction site. substrate site. inhibitor site. 0 1644446423 MultipleChoice 8
0 1644446425 MultipleChoice 10
primary structure. secondary structure.
3 3 Lactose 0 1644446433 MultipleChoice 23
linear pyramidal bent tetrahedral None of the above 0 1644446434 MultipleChoice 27
Br 2 CH 4 NF 3 0 1644446437 MultipleChoice 28
acid?
None of the above 0 1644446439 MultipleChoice 31
4 1
(^1 )
2 3
0 1644446445 Matching 5
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 :
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
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