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CHEM 120 Final exam with

100% verified answers-2024-

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 = t

+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 P

(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)

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)

0 1644446450 Short 1 molarity = moles solute / liters solution 0.25 M = moles NaOH / 0.035 L moles NaOH = 0. moles NaOH 3.5 pints is equivalent to 1656.

3. (TCO^ 3)^ What^ is^ the^ name^ of^ the^ following^ compound:^ Zn 3 P 2?^ (Points^ : 5)

0 1644446452 Short 6

3

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

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)

1. (^) (TCO 7) (a, 5 pts) Given that the molar mass of H 3 PO 4 is 97.994 grams, determine the number of grams of H 3 PO 4 needed to prepare 0.75L of a 0.25M H 3 PO 4 solution. Show your work. (b, 5 pts) What volume, in Liters, of a 0.25 M H 3 PO 4 solution can be prepared by diluting 50

mL of a 2.5M H 3 PO 4 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. grams H3PO b. M1V1 = M2V 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 NaO Volume = 43 g * (1 mL/2.13 g) = 20.19 mL H 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: It's Zinc Phosphide 0 1644446453 Short 7 Silver nitrate 0 1644446454 Short 11 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. 0 1644446456 Short 14

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 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 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 (10 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 : 20)

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 1644446471 Essay 21

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 (10 pts)?

(Points : 20)

1. (TCO 12) Transcription is the process by which DNA passes information to (Points : 5)

another strand of DNA. Grams: 1.75 mol cl2*2/133.34=0.026g Molar Mass: Al = 26.981 g/molCl = 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 Balance: 2Al + 3 Cl2 > 2 AlCl Moles: 1.75 mol Cl2 * (2 mol AlCl3 / 3 mol Cl2) = 1.17 mol AlCl 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

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.

tertiary structure. primary and secondary structures. secondary and tertiary structures. 0 1644446427 MultipleChoice 15

5. (TCO 12) Which of the following is a disaccharide? (Points : 5)

Cellulose Starch Fructose Galactos e Lactose 0 1644446429 MultipleChoice 17

6. (TCO 12) In cells, protein synthesis occurs on the (Points : 5)

Golgi apparatus. mitochondria. nucleus. cytoplasm. ribosomes. 0 1644446431 MultipleChoice 20

7. (TCO 12) Which of the following is a monosaccharide? (Points : 5)

Cellulose Starch Galactos e

Mannose

3 3 Lactose 0 1644446433 MultipleChoice 23

8. (TCO 4) The best description of the shape of an ammonia molecule is. (Points : 5)

linear pyramida l bent tetrahedr al 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

3 3

11. (TCO 8) According to the Bronsted-Lowry definition, which chemical in the following

reaction is the base?

H 2 CO 3 + H 2 O -> H 3 O+ + HCO - (Points : 5)

H 2 O

H 2 CO 3

HCO -

H 3 O+

None of the above 0 1644446442 MultipleChoice 34

12. (TCO 2) Match the atom with the appropriate description.

(Points : 15)

Potential Matches:

1 : Boron (B)

2 : Carbon

(C) 3 : Neon

(Ne)

4 : Nitrogen (N)

5 : Lithium (Li)

Answer

: An atom, which has a mass number of 7, with 4 neutrons

: An atom, which has a mass number of 11, with 6 neutrons

: An atom, which has a mass number of 12, with 6 neutrons

: An atom, which has a mass number of 14, with 7 neutrons

: An atom, which has a mass number of 20, with 10 neutrons

0 1644446444 Matching 1

13. (TCO 10) Match the organic compound with its use or characteristic.

(Points : 15)

Potential Matches:

1 : Disinfectant

2 : Fingernail polish remover

3 : Alkaloid

4 : Insecticide

(^5 ) (^1 ) (^2 ) (^4 ) (^3 )

5 : Pineapple flavor

Answer

: Dichlorodiphenyltrichloroethane

4 1

:

Formaldehyd e (^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 ) (^5 ) (^1 ) (^4 ) (^2 ) (^5 ) (^3 )

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.

instruments and appliances, and it is w idely used for home insulation.

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) mass of 180 becomes 176 atomic number of 74 becomes 72 name: Hafnium Symbol: Hf 6. (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)? (Points : 20) 1. The prevalent plastic polyethylene is the simplest and least expensive synthetic polymer. It is familiar today in the plastic bags used for packaging fruit and vegetables, in garment bags for dry-cleaned clothing, in garbage-can liners, and in many other items. Polyethylene is made from ethylene (CH2 = CH2) an unsaturated hydrocarbon (Chapter 9) produced in large quantities from the cracking of petroleum, a process by w hich large hydrocarbon molecules are broken dow n into simpler hydrocarbons. 2. Polypropylene is a tough plastic material that resists moisture, oils, and solvents. It is molded into hard-shell luggage, battery cases, and various kinds of appliance parts. It is also used to make packaging material, fibers for textiles such as upholstery fabrics and carpets, and ropes that float. Because of its high melting point (121 °C), polypropylene objects can be sterilized w ith steam 3. Polystyrene is the plastic used to make transparent disposable drinking cups. With color and filler added, it is the material of thousands of inexpensive toys and household items. When a gas is blow n into polystyrene liquid, it foams and hardens into the familiar material of ice

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 : (^) 20) 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. g/mol (1 x Al) + (3 x Cl) = (1 x 26.981g/mol) + ( 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 (10 pts)?

(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) molarity = moles solute / liters solution 0.75 M = moles NaOH / 0.045 L moles NaOH = 0. moles NaOH

0 1645236738 Short 4

3. (TCO^ 3)^ What^ is^ the^ name^ of^ the^ following^ compound:^ Na 2 S?^ (Points^ :^ 5) 0 1645236739 Short 8 4. (TCO^ 3)^ What^ is^ the^ name^ of^ the^ following^ compound:^ AgNO 3?^ (Points^ :^ 5) 0 1645236740 Short 11 5. (TCO^ 6) Calculate^ the^ volume,^ in liters,^ of^ a 4.12^ mol H 2 S(g) at^ 60 degrees^ C and 1.75^ atm. (Points : 5) 6. (TCO 6) A gas at a temperature of 185 degrees C occupies a volume of 575 mL. Assuming constant pressure, determine the volume at 15 degrees C. Show your work. (Points : 5) 7. (TCO 6) A sample of helium gas occupies 1045 mL at 721 mmHg. For a gas sample at constant temperature, determine the volume of helium at 745 mmHg. Show your work. (Points : 5) 1 Inch = 0.0254 Meter 175*0.0254= 4.445 meters Sodium sulfide Silver nitrate Given that n = 4.12 mol; and temperature is 60 degrees, w hich is 333 K; P = 1.75 atm; w e w ant to calculate volume in liters. First, convert your temperatures to Kelvin: T = t + 273. Place the know n values in this equation: PV= nRT. So, V = nRT/P = Using Charles’ Law , (V1/T1) = (V2/T2). First, convert temperature to KELVIN (T1 = t1 +273) Thus, T1 = 185 + 273 =

458. We have V1 (575 mL) & T2 = (15 + 273) = 288. V2 = (V1T2)/T1 = ( mL288)/458 = 361.57 mL.

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 Using Boyle’s law , P1V1 = P2V2. We have V1 (1045 mL), P (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. TGCAGTACCG

CH 3 CH=CHCH 2 CH=CHCOOH

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 Galactos e 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) Mannos e Starch Cellulos e Glucose Lactose 8. (TCO 4) The best description of the shape of an ammonia molecule is. (Points : 5) linear

pyramid al

4 4 2 3 3 2 bent tetrahedr al None of the above

9. (TCO 4) Which of the following substances does not have polar covalent bonds? (Points : 5) CO 2 Br 2 CH 4 NF 3 10. (TCO 8) According to the Bronsted-Lowry definition, which chemical in the following reaction is the acid? NH +^ + H O -> H O+^ + NH (Points : 5) H 3 O + NH + NH 3 H 2 O None of the above 11. (TCO 8) According to the Bronsted-Lowry definition, which chemical in the following reaction is the base? HCN + NO -^ -> CN-^ + HNO (Points : 5) HNO 2 NO 2 - HCN CN- None of the above 12. (TCO 2) Match the atom with the appropriate description. (Points : 15) Potential Matches: 1 : Nitrogen (N) 2 : Carbon (C)

3 : Boron (B) 4 : Fluorine (F) 5 : Beryllium (Be) Answer

: An atom, which has a mass number of 9, with 5 neutrons : An atom, which has a mass number of 11, with 6 neutrons : An atom, which has a mass number of 12, with 6 neutrons : An atom, which has a mass number of 19, with 10 neutrons : An atom, which has a mass number of 14, with 7 neutrons 0 1645236734 Matching 2

13. (TCO 10) Match the organic compound with its use or characteristic. (Points : 15) Potential Matches: 1 : Component of ant sting 2 : Found in antifreeze 3 : Orange flavor 4 : Propellants 5 : Fishy odor Answer : Octyl acetate : CFCs : Formic acid : Ethylene glycol : Trimethylamine 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 (^5 1) (^3 2) (^2 3) (^4 4) (^1 5) (^3 1) (^4 2) (^1 3) (^2 4) (^5 5) (^1 1) (^4 2) (^2 3) (^5 4) (^3 5)