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CHEM 104 Chemistry II Portage Learning Exam 1 Questions and Verified Answers (2025l2026) This document contains the complete set of verified questions and answers for Exam 1 of CHEM 104 (Chemistry II) from Portage Learning. It covers key topics typically included in the first exam, such as intermolecular forces, colligative properties, reaction kinetics, and chemical equilibrium. The answers are reviewed and confirmed for accuracy, making it an ideal study resource for exam preparation.
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Not yet graded / 10 pts le.pdf) Click this link to access the Periodic Table. (https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_Tab
This may be helpful throughout the exam. In the reaction of gaseous N 2 O 5 to yield NO 2 gas and O 2 gas as shownbelow, the following data table is obtained: 2 N 2 O5 (g) → 4 NO2 (g) + O2 (g) Data Table # Time (sec) [N 2 O 5 ] [O 2 ] 0 0.200 M 0 300 0.182 M 0.009 M 600 0.166 M 0.017 M 900 0.152 M 0.024 M 1200 0.140 M 0.030 M 1800 0.122 M 0.039 M
Not yet graded / 10 pts le.pdf) Click this link to access the Periodic Table. (https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_Tab
This may be helpful throughout the exam. The following rate data was obtained for the hypothetical reaction: C + D → X + Y Experiment # [C] [D] rate 1 0.50 0.50 3. 2 1.00 0.50 6. 3 1.00 1.00 48.
Click this link to access the Periodic Table. (https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_Table.pdf) This may be helpful throughout the exam. ln [A] - ln [A] 0 = - k t 0.693 = k t1/ An ancient sample of cloth was found to contain 21.4 % 14 C content as compared to a present-day sample. The t1/2 for 14 C is 5720 yrs. Show the calculation of the decay constant (k) and the age of the cloth. Your Answer: K= 0.693/t1/2=0.693/5720 = 1.2115x10^(-4) ln (21.4) - ln (100) = - kt 3.06-4.61 = 1.55 = (1.2115x10^(-4))t t= 12794 years Not yet graded / 10 pts
0.693 = k t1/ = 1.2115 x 10 -^4 0.693 = k (5720 yrs) k = 0.693 / 5720 ln [A] - ln [A] 0 = - k t = - (1.2115 x 10-^4 ) t 3.0634 - 4.6052 = - (1.2115 x 10-^4 ) t / - 1.2115 x 10-^4 = 12,726 yrs ln 21.4 - ln 100 t (age) = - 1.
Your Answer: Reaction is exothermic because it lost energy. The products are at a lower energy level than the reactants. The reaction is not spontaneous because of Eact is (large). Large Eact = nonspontaneous ∆H-^ =^ exothermic
Click this link to access the Periodic Table. (https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_Table.pdf) This may be helpful throughout the exam. Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.300 mole of CO and 0.900 mole of H 2 in a 3.00 liter container forms an equilibrium mixture containing 0.117 mole of H 2 O and corresponding amounts of CO, H 2 , and CH 4. Not yet graded / 10 pts
At equilibrium H 2 O = 0.117 mole (as stated) CH 4 = 0.117 mole (1 mole of CH 4 forms for every mole of H 2 O that is formed) CO = 0.300 - 0.117 mole (1 mole of CO reacts for every mole of H 2 O that is formed) H 2 = 0.900 - 3 x 0.117 mole (3 mole of H 2 reacts for every mole of H 2 O that is formed) Change all amounts to moles/L before entering in Kc expression: H 2 O = 0.117 mole / 3.00 L = 0.0390 M CH 4 = 0.117 mole / 3.00 L = 0.0390 M CO = 0.183 mole / 3.00 L = 0.0610 M H 2 = 0.549 mole / 3.00 L = 0.183 M Kc = [CH 4 ] [H 2 O] = [0.0390] [0.0390] = 4. [CO] [H 2 ]^3 [0.0610] [0.183]^3
Why is a molecularity greater than 3 unheard of? Your Answer: This is a collision reaction with reactant molecules. There cant be more than 3 molecules that may collide with each other for an effective collision. Not yet graded / 10 pts
This may be helpful throughout the exam.
The equilibrium reaction below has the following equilibrium mixture concentrations: 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 H 2 O at equilibrium is increased to [0.100], 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 H 2 (g) CH 4 (g) + H 2 O (^) (g) Your Answer: Since H2O is one of the products and the concentration of H2O is increasing, the reaction shifts to the left to obtain equillibrium Q= [H2O] x[CH4]/ [CO] x [H2]^ (.1)x(0.0380)/ ([0.620) x [0.186])
Since Q is greater than Kc reaction shifts to the left Kc = 3.62 when H 2 O = [0.0380], CH 4 = [0.0380], CO = [0.0620] andH 2 = [0. M] When H 2 O = [0.100], Q = [0.100] [0.0380] = 9. [0.0620] [0.186]^3 The reaction must shift briefly in the reverse direction to decrease the [H 2 O] to come back to equilibrium.This is in agreement with Qc
Kc which also predicts the reaction will proceed to the left.
Click this link to access the Periodic Table. (https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_Table.pdf) This may be helpful throughout the exam. The equilibrium reaction below has the Kc = 3.93. If the volume of the system at equilibrium is decreased from 6.00 liters to 3.00 liters, how andfor what reason will the equilibrium shift? Be sure to calculate the value ofthe reaction quotient, Q, and use this to confirm your answer. CO (^) (g) + 3 H 2 (g) CH 4 (g) + H 2 O (^) (g) Your Answer: 4 mole --> 2 mole this reaction favors volume reduction. since volume is decreasing theforward reaction is favored. The equillibrium shifts to the right The new volume of Kc is equal to 1/2Q=
Q< Kc the reaction shifts to the right
When volume decreases from 6.00 to 3.00, the pressure doubles and the concentration of all gases (CO, H 2 , CH 4 , and H 2 O) doublesso: (at equilibrium) Qc =Kc = [CH 4 ] [H 2 O] = 3. [CO] [H 2 ]^3 (volume halved = pressure doubled = conc doubled) Qc = [2 CH 4 ][2 H 2 O] = Kc [2 H 2 ]^3
The reaction must shift briefly in the direction that decreases the pressure by going toward the side with the lesser moles of gas (forward direction : 4 moles of gas yields 2 moles of gas) to comeback to equilibrium.This is in agreement with Qc < Kc: the reactionwill proceed to the right (in the direction of the products).
Not yet graded / 10 pts le.pdf) Click this link to access the Periodic Table. (https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_Tab
