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Portage learning CHEM 104 Module 1 to 6 Exam Questions and Answers (2025 / 2026), Exams of Chemistry

Geneva CollegeChemistry

Portage learning CHEM 104 Module 1 to 6 Exam Questions and Answers (2025 / 2026) (Verified Answers). Chem 104 final exam questions, Chem 104 final exam pdf, Chem 104 final exam answer key, Chem 104 final exam answers, CHEM 104 final, Portage chem 104 exam 4, Portage Learning CHEM 104 Lab 2, Portage learning chem 104 module 3 problem set, Chem 104 lab pdf Chem 104 lab answers, Chem 104 lab questions, Chem 104 module 1 module 6 exam pdf, Chem 104 module 1 module 6 exam answer key, Chem 104 module 1 module 6 exam answers, Portage Learning CHEM 104 final exam, Portage learning chem 104 final exam questions, Portage learning chem 104 final exam pdf, Portage learning chem 104 final exam answer key, Portage learning chem 104 final exam answers, Portage Learning Chemistry 104 Answers, Portage learning chem 104 pdf Portage learning chem 104 answers, Portage learning chem 104 answer key, Portage Learning CHEM 104 Lab exam 1, Portage Learning Chemistry 104 Answers

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CHEM 104
MODULE 1 , 2, 3, 4, 5, 6, EXAM COMPILED
TABLE OF CONTENTS
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CHEM 104 MODULE 5……………………………………………
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Download Portage learning CHEM 104 Module 1 to 6 Exam Questions and Answers (2025 / 2026) and more Exams Chemistry in PDF only on Docsity!

CHEM 104

MODULE 1, 2, 3, 4, 5, 6, EXAM COMPILED

TABLE OF CONTENTS

CHEM 104 MODULE 1……………………………………………

CHEM 104 MODULE 2……………………………………………

CHEM 104 MODULE 3……………………………………………

CHEM 104 MODULE 4……………………………………………

CHEM 104 MODULE 5……………………………………………

CHEM 104 MODULE 6……………………………………………

  1. rate = (0.014 - 0) / (300 - 0) = 4.67 x 10

    mol/Ls

  2. rate = (0.090 - 0.083) / (3000 - 2400) = 1.167 x 10

    mol/Ls
  3. The late instantaneous rate is smaller than the early instantaneous rate.

Question 2 The following rate data was obtained for the hypothetical reaction: A + B → X + Y Experiment # [A] [B] rat e 1 0.50 0.50 2. 2 1.00 0.50 8. 3 1.00 1.00 64. 0

  1. Determine the reaction order with respect to [A].
  2. Determine the reaction order with respect to [B].
  3. Write the rate law in the form rate = k [A] n [B] m (filling in the correct exponents).
  4. Show the calculation of the rate constant, k. Your Answer: rate = k [A]x [B]y rate 1 / rate 2 = k [0.50]x^ [0.50]y^ / k [1.00]x^ [0.50]y^ 2.0 / 8.0 = [0.50]x^ / [1.00]x 0.25 = 0.5x x = 2 rate 2 / rate 3 = k [1.00]x^ [0.50]y^ / k [1.00]x^ [1.00]y^ 8.0 / 64.0 = [0.50]y^ / [1.00] y 0.125 = 0.5y^ y = 3 rate = k [A]^2 [B]^3 2.0 = k [0.50] 2 [0.50] 3 k = 64 Question 3 ln [A] - ln [A] 0 = - k t 0.693 = k t1/ An ancient sample of paper was found to contain 19.8 % 14 C content as

Your Answer: 0.693 = k t1/ 0.693 = k (5720) k = 1.21 x 10

ln [A] - ln [A] 0 = - k t ln 19.8 - ln 100 = - 1.21 x 10-4^ t t = 13, 384 years Question 4 Using the potential energy diagram below, state whether the reaction described by the diagram is endothermic or exothermic and spontaneous or nonspontaneous, being sure to explain your answer. Your Answer: The reaction is exothermic since it has a negative heat of reaction and it is nonspontaneous because it has relatively large Eact. Question 5 Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H 2 in a 8.00 liter container forms an equilibrium mixture containing 0.309 mole of H 2 O and corresponding amounts of CO, H 2 , and CH 4. CO (^) (g) + 3 H2 (g) CH 4 (g) + H 2 O (^) (g) Your Answer: 0.309 mole of H 2 O formed = 0.309 mole of CH 4 formed

0.309 mole of H 2 O formed = 0.800 - 0.309 = 0.491 mole CO 0.309 mole of H 2 O formed = 3 x 0.309 mole H 2 reacted = 2.40 - (3 x 0.309) = 1.473 mole H 2 [CO] = 0.491 mole / 8.00 L = 6.1375 x 10

  • M [H2] = 1.473 mole / 8.00 L = 18.4125 x 10 -2^ M [CH4] = 0.309 mole / 8.00 L = 3.8625 x 10 -2^ M [H2O] = 0.309 mole / 8.00 L = 3.8625 x 10 -2^ M Kc = [3.8625 x 10-2] [3.8625 x 10 -2] / [6.1375 x 10-2] [18.4125 x 10 -2]^3 Kc = 3. Question 6 Explain the terms substrate and active site in regard to an enzyme. Your Answer: Enzymes are large protein molecules that act as catalysts and increases the rate of a reaction. The catalysts act only on one type of substance to cause one type of reaction and this is called a substrate. Active sites are enzymes that are spherical in shape together ith a group of atoms on a surface of protein. These active sites bind with the substrate causing it to undergo a reaction. The substrate and active site forms a complex that creates a new pathway with lower activation energy and thus speed up a reaction. Question 7 The reaction below has the indicated equilibrium constant. Is the equilibrium mixture made up of predominately reactants, predominately products or significant amounts of both products and reactants. Be sure to explain your answer. 2 O 3 (g) 3 O 2 (g) Kc = 2.54 x 10^12 Your Answer: Kc = 2.54 x 1012 is very large. Thus, the equilibrium mixture is made up of predominantly products.

H 2 O = [0.0380], CH 4 = [0.0380], CO = [0.0620] and H 2 = [0.186] with Kc = 3.62. If the concentration of CO at equilibrium is increased to [0.200], how and for what reason will the equilibrium shift? Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer. Your Answer: Kc = 3. CO (^) (g) + 3 H2 (g) CH 4 (g) + H 2 O (^) (g) Qc = [0.200] [0.0380] / [0.0620] [0.186]^3 Qc = 19. Qc is greater than Kc and thus the equilibrium will shift to the left towards the direction of the reactants. Kc = 3.62 when H 2 O = [0.0380], CH 4 = [0.0380], CO = [0.0620] and H 2 = [0. M] When CO = [0.200], Q = [0.0380] [0.0380] = 1. [0.200] [0.186]^3 The reaction must shift briefly in the forward direction to decrease the [CO] to come back to equilibrium.This is in agreement with Qc < Kc which also predicts the reaction will proceed to the right. new CO concentration is 0. Question 9 The equilibrium reaction below has the Kc = 3.93. If the volume of the system at equilibrium is decreased from 6.00 liters to 2.00 liters, how and for what reason will the equilibrium shift? Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer. CO (^) (g) + 3 H2 (g) CH 4 (g) + H 2 O (^) (g) Your Answer: Initial volume = 6. L Final volume = 2.00 L Volume is

reduced by 3. Pressure is inversely proportional to volume, thus pressure is tripled and concentration of gases is also tripled. Final concentration = Initial concentration x 3

and the concentration of the reactants decreases and thus increases the value of Kc.

4 4 4 4 Module 2: Question 1 Identify each of the compounds below as ACID, BASE or SALT on the basis of their formula and explain your answer. (1) Cr(OH) 3 (2) HAsO 4 (3) CoCO 3 Your Answer:

  1. base - contains Cr metal + OH polyatomic group
  2. acid - contains H + polyatomic group (AsO 4 )
  3. salt - contains Co metal + polyatomic group (CO 3 ) Question 2 For the Brønsted-Lowry acid base reactions shown below, list the stronger acid, stronger base, weaker acid and weaker base in the answer blanks provided: NH
      H 2 PO -

NH 3 + H 3 PO 4

Stronger acid: Stronger base: Weaker acid: Weaker base: Your Answer: Stronger acid: H 3 PO 4 Stronger base: NH 3 Weaker acid: NH + Weaker base: H 2 PO

Question 3

4 HClO 4 or HBrO 4 or HIO 4 Stronger acid is: Explanation: Formula of conjugate base of the stronger acid: Your Answer: Stronger acid is: HClO 4 > HBrO 4 > HIO 4 Cl is more electronegative than Br so HClO 4 is stronger acid than HBrO 4 and so is Br to I, Br is more electronegative than I so HBrO 4 is stronger than HIO 4. Stronger acid is: HClO 4 Explanation: Cl has the higher electronegativity (of Cl, Br or I) which makes the H-O bond of HClO 4 most polar and most likely to form H+ Formula of conjugate base of stronger acid: conjugate base of HClO 4 is ClO - Question 4 Show the calculation of the [H+] and pH of a 0.00350 M solution of the strong acid H 2 SO 4. Your Answer: [H+] = 2 x [H 2 SO 4 ] = 2 x (0.00350] [H+] = = 0.007 M pH = - log [H+] = - log ( 0.007) pH = 2. Question 5 In the titration of 15.0 mL of H 2 SO 4 of unknown concentration, the phenolphthalein indicator present in the colorless solution turns pink when 26. mL of 0.130 M NaOH is added. Show the calculation of the molarity of the H 2 SO 4.

2 2 2 Your Answer: (Ma x mLa)/1000 x Sa/Sb = (Mb x mLb)/1000 (Ma x 15.0)/1000 x 1/ = (0.130 x 26.4)/1000 MH2SO4 = 0.458 M (H 2 SO 4 is the acid and NaOH is the base) Ma x mLa / 1000 x Sa / Sb = Mb x mLb / 1000 Ma x 15.0 / 1000 x 2 / 1 = 0.130 x 26.4 / 1000 MH2SO4 = 0.114 M ratio: 2 / 1 Question 6 Show the calculation of the [H+], pH and % ionization for 0.645 M acetic acid (HC 2 H 3 O 2 ) HC 2 H 3 O 2 H+^ + C 2 H 3 O

  • Ka = 1.8 x 10 - Your Answer: HC 2 H 3 O 2 H+^ + C 2 H 3 O - 0.645 0 0 -x +x + x 0.645-x x x Ka = 1.8 x 10

Ka = [C 2 H 3 O

] [H

+] / [HC

2 H 3 O 2 ]

1.8 x 10-5^ = (x) (x) / (0.645-x) x^2 = 1.8 x 10-

2 2 3 2 4 4 pH = - log [H

] = - log (3.41x

  • ) pH = 2. % ionization = ([C 2 H 3 O - ]/ [HC H O ]) x 100 % ionization = (3.41x10-3^ / 0.645) x 100 % ionization = 0.53% Question 7 Predict and explain whether a solution of LiF is acidic, basic or neutral. Your Answer: Basic since Li hydrolyzes to form strong base [LiOH]. LiF: Basic since F-^ hydrolyzes to form a weak acid (HF) and OH- Question 8 Show calculation of the pH of a buffer prepared by mixing 0.100 M NH 4 Cl and 0.0750 M NH 3. NH 3 + H 2 O (^) (liq) NH +^ + OH- Ka = 1.8 x 10

Your Answer: NH 3 + H 2 O (^) (liq) NH +^ + OH- 0.0750 0.100 0 -x +x + x 0.0750-x Ka = 1.8 x 10 - 0.100+x x

Ka = [NH 4

] [OH ] / [NH 3 ]

1.8 x 10 -5^ = (0.100+x) (x) / (0.0750-x)