A 1.00 mol of helium, He gas fills up an empty balloon to a volume of 1.50 L. Calculate the volume of the balloon if an additional 2.50 mol of He gas is added. The temperature and pressure of the gas are constant. 1

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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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A 1.00 mol of helium, He gas fills up an empty balloon to a volume of 1.50 L.
Calculate the volume of the balloon if an additional 2.50 mol of He gas is added. The
temperature and pressure of the gas are constant.
1
[2]
A 4.00 L of methane, CH4 is burned in excess oxygen at 25.0 °C and 1.50 atm to
produce water and carbon dioxide.
2
CH4 (g) + 2 02(g) → CO2(g) + 2 H20(1)
Calculate the volume of carbon dioxide produced after the reaction has stopped.
[Assume the temperature and pressure of product are the same with reactant.
[6]
R=0.0821
atm. LmolK1]
3
(a)
State Dalton's Law.
[1]
(b) A 90.0 mL of neon, Ne at 27.0 °C and 5.00 atm is mixed with nitrogen, N2 and
transferred into a 200.0 mL container at the same temperature. The total pressure
of the mixture is 4.25 atm. Calculate the partial pressure of Ne and N2.
[4]
4
State two basic assumptions of kinetic molecular theory of gasses. [2]
Transcribed Image Text:A 1.00 mol of helium, He gas fills up an empty balloon to a volume of 1.50 L. Calculate the volume of the balloon if an additional 2.50 mol of He gas is added. The temperature and pressure of the gas are constant. 1 [2] A 4.00 L of methane, CH4 is burned in excess oxygen at 25.0 °C and 1.50 atm to produce water and carbon dioxide. 2 CH4 (g) + 2 02(g) → CO2(g) + 2 H20(1) Calculate the volume of carbon dioxide produced after the reaction has stopped. [Assume the temperature and pressure of product are the same with reactant. [6] R=0.0821 atm. LmolK1] 3 (a) State Dalton's Law. [1] (b) A 90.0 mL of neon, Ne at 27.0 °C and 5.00 atm is mixed with nitrogen, N2 and transferred into a 200.0 mL container at the same temperature. The total pressure of the mixture is 4.25 atm. Calculate the partial pressure of Ne and N2. [4] 4 State two basic assumptions of kinetic molecular theory of gasses. [2]
For the reaction X + Y → 2Z, the following data were obtained:
Experiment
[X] (M)
[Y] (M)
Initial Rate (M/s)
8.00 x 10-4
8.00 x 10-4
5.90 x 10-3
1.20 x 10-2
5.60 x 10-8
2.24 x 10-7
1
3
1.60 x 10-3
5.90 x 10-3
1.10 x 10-7
4
2.30 x 10-3
1.30 x 10-2
8.90 х 10-7
(a)
Based on the data above, calculate the rate law.
[6]
(b)
Calculate the rate constant of the reaction using experiment 1.
[3]
The rate constant for decomposition of N205 is 6.2 x 10-4 min-1. Calculate the time it takes
for the concentration of N205 to decrease to 16.5% from its initial concentration. The
initial concentration of N2Os is 1.00 M.
[2]
7 State two criteria for a proposed mechanism to be a valid.
[2]
8
Given the following reaction mechanism:
[2]
Step 1: NO2CI (g) → NO2 (g) + CI (g)
Step 2: NO2CI (g) + CI (g)
(slow)
(fast)
→ NO2 (g) + Cl2 (g)
(a)
Write the rate law for the rate determining step.
[1]
(b)
Identify the intermediate for the reactions above.
[1]
Transcribed Image Text:For the reaction X + Y → 2Z, the following data were obtained: Experiment [X] (M) [Y] (M) Initial Rate (M/s) 8.00 x 10-4 8.00 x 10-4 5.90 x 10-3 1.20 x 10-2 5.60 x 10-8 2.24 x 10-7 1 3 1.60 x 10-3 5.90 x 10-3 1.10 x 10-7 4 2.30 x 10-3 1.30 x 10-2 8.90 х 10-7 (a) Based on the data above, calculate the rate law. [6] (b) Calculate the rate constant of the reaction using experiment 1. [3] The rate constant for decomposition of N205 is 6.2 x 10-4 min-1. Calculate the time it takes for the concentration of N205 to decrease to 16.5% from its initial concentration. The initial concentration of N2Os is 1.00 M. [2] 7 State two criteria for a proposed mechanism to be a valid. [2] 8 Given the following reaction mechanism: [2] Step 1: NO2CI (g) → NO2 (g) + CI (g) Step 2: NO2CI (g) + CI (g) (slow) (fast) → NO2 (g) + Cl2 (g) (a) Write the rate law for the rate determining step. [1] (b) Identify the intermediate for the reactions above. [1]
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