CHEM120 Week 8 Final Exam-With 100% Verified Solutions Latest Update with Complete Solutio, Exams of Nursing

Download CHEM120 Week 8 Final Exam-With 100% Verified Solutions Latest Update with Complete Solutio and more Exams Nursing in PDF only on Docsity! 3 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 1656.116 1 pint = 473.176 ml 3.5 pints* 473.176mL = 1656.116mL It's Zinc Phosphide Silver nitrate 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. CHEM120 Week 8 Final Exam-With 100% Verified Solutions Latest Update with Complete Solutions 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) 0 1644446450 Short 1 2. (TCO 1) How many mL are in 3.5 pints? Show your work. (Points : 5) 0 1644446452 Short 6 3. (TCO 3) What is the name of the following compound: Zn3P2? (Points : 5) 0 1644446453 Short 7 4. (TCO 3) What is the name of the following compound: AgNO3? (Points : 5) 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) 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. 0 1644446470 Essay 16 7. (TCO 5) Given the following unbalanced chemical equation: Al + Cl2-> AlCl3 (a, 5 pts) Balance the equation. (b, 5 pts) How many moles of AlCl3 are produced from 1.75 mole of Cl2? Show your work. (c, 5 pts) What is the molar mass of AlCl3? Show your work. (d, 5 pts) Calculate the number of grams of AlCl3 produced from 1.75 mol Cl2. Show your work. (Points : 20) Balance: 2Al + 3 Cl2 > 2 AlCl3 Moles: 1.75 mol Cl2 * (2 mol AlCl3 / 3 mol Cl2) = 1.17 mol AlCl3 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 Grams: 1.75 mol cl2*2/133.34=0.026g 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) AAACGTGTGCTAACA w ill become UUUGCACACGAUUGU Peptide sequence is UUU GCA CAC GAU UGU w hich is Phe-Ala-His-Asp-Cys 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 12) Transcription is the process by which DNA passes information to (Points : 5) another strand of DNA. 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) CH3CH2CH2CH=CHCOOH CH3CH=CHCH=CHCOOH CH3CH2CH2CH2CH2COOH CH3CH2CH2CH2CH2CH=CHCOOH CH3CH=CHCH2CH=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 Galactose 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 Galactose Mannose Lactose 3 : Ethylamine 4 : Heptanal 5 : Pentyl acetate Answer : CH3COOH : CH3CH2CH2CH2CH2CH2CHO : CH3CH2CH2OH : CH3COOCH2CH2CH2CH2CH3 : CH3CH2NH2 Week 8 : Final Exam - Final Exam 6870487 http://takeexam.n en-US 00:51:58 Time Remaining: Page: 1 2 3 CHEM120 Final Exam Page 3 31910153 20733303 3 rldbqn=1 MultipleSections NavigateFreely False 0 0 /main/CourseMod 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 0.25L 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. (Points : 10) en-US 1 4 2 5 3 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 0 1645236746 Essay 1 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) 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. 0 1645236747 Essay 27 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) instruments and appliances, and it is w idely used for home insulation. 0 1645236748 Essay 8 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 0 1645236749 Essay 11 5. (TCO 1) Recently, you’ve noticed that your 10-year-old vacuum cleaner is not picking up the animal fur from your carpet as it has in the past. You have decided to invest in a new vacuum cleaner. Before you make a decision on which model to purchase from manufacturer XYZ, your neighbors offer to let you use their vacuums to help you in your decision-making process. Neighbor A has a vacuum made by manufacturer XYZ and is 5 years old. Neighbor B has a vacuum made by manufacturer XYZ and is a new model that has just been released. Formulate a testable hypothesis (5 pts). Design an experiment to test your hypothesis, and include your controls (5 pts). Be specific in your experimental design. Predict the experimental results that would support your hypothesis (5 pts). (Points : 15) 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 chests and disposable coffee cups. The polymer can easily be formed into shapes as packing material for shipping Week 8 : Final Exam - Final Exam 6870487 http://takeexam.n en-US 00:51:00 Time Remaining: Page: 1 2 3 CHEM120 Final Exam Page 2 31910152 20733303 2 rldbqn=1 MultipleSections NavigateFreely False /main/CourseMod 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) molarity = moles solute / liters solution 0.75 M = moles NaOH / 0.045 L moles NaOH = 0.03375 moles NaOH 0 1645236737 Short 2 2. (TCO 1) How many meters are in 175 inches? Show your work. (Points : 5) 1 Inch = 0.0254 Meter 175*0.0254= 4.445 meters 0 1645236738 Short 4 3. (TCO 3) What is the name of the following compound: Na2S? (Points : 5) 00:51:58 en-US Sodium sulfide 0 1645236739 Short 8 4. (TCO 3) What is the name of the following compound: AgNO3? (Points : 5) Silver nitrate 0 1645236740 Short 11 5. (TCO 6) Calculate the volume, in liters, of a 4.12 mol H2S(g) at 60 degrees C and 1.75 atm. (Points : 5) nRT/P = 0 1645236741 Short 13 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) 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 = (V1*T2)/T1 = (575 mL*288)/458 = 361.57 mL. 0 1645236742 Short 17 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) Using Boyle’s law , P1V1 = P2V2. We have V1 (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. 0 1645236743 Short 20 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) 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 = 00:51:00 TGCAGTACCG 0 1645236744 Short 24 Week 8 : Final Exam - Final Exam 6870487 http://takeexam.n en-US 00:50:31 Time Remaining: Page: 1 2 3 CHEM120 Final Exam Page 1 31910149 20733303 1 rldbqn=1 MultipleSections NavigateFreely False /main/CourseMod 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 0 1645236716 MultipleChoice 3 en-US H2O None of the above 0 1645236731 MultipleChoice 32 11. (TCO 8) According to the Bronsted-Lowry definition, which chemical in the following reaction is the base? HCN + NO2- -> CN- + HNO2 (Points : 5) HNO2 NO2 - HCN CN- None of the above 0 1645236732 MultipleChoice 36 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 5 1 3 2 2 3 4 4 1 5 3 1 4 2 00:50:31 : Formic acid : Ethylene glycol : Trimethylamine 0 1645236735 Matching 6 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 : CH3COOH : CH3CH2CH2CH2CH2CH2CHO : CH3CH2CH2OH : CH3COOCH2CH2CH2CH2CH3 : CH3CH2NH2 0 1645236736 Matching 9 1 3 2 4 5 5 1 1 4 2 2 3 5 4 3 5