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julsineya [31]
3 years ago
7

You are interested in how different air temperatures affect the speed at which gas molecules move. With a classmate, you have pl

anned the following investigation:
Choose a very hot day in the classroom.

The students close all of the windows and doors in the classroom and turn off any fans.
The students measure the temperature in the classroom with a thermometer.
One student stands in the corner of the classroom with a stopwatch.
Another student, 10 meters away in the opposite corner of the room, opens a container filled with a safe, but strong-smelling gas.
The student with the stopwatch measures the amount of time before they can smell the gas.
Repeat the steps above on a very cold day in the classroom.
Identify the independent variable in this investigation from the list below:

A. The temperature of the room
B. Amount of time before they smell the gas
C. The distance between the two students in the room
D. Not enough information is provided to identify the independent variable
E. The type of gas in the container
Chemistry
2 answers:
Andru [333]3 years ago
4 0
C I beLive (I’m not positive)
Anna35 [415]3 years ago
3 0
It’s A, The temperature of the room.
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How many moles of water will be produced when 3 moles of ethane are burned
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The reaction of ethane burning:
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The following mechanism has been proposed for the gas-phase reaction of chloroform (CHCl3) and chlorine. Cl2 ⇌ 2Cl (fast, revers
SVEN [57.7K]

Answer:

Explanation:

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Q&A Library

The following mechanism has been proposed for the gas-phase reaction of chloroform (CHCl3) and chlorine.Cl2 ⇌ 2Cl (fast, reversible) Cl + CHCl3 → HCl + CCl3 (slow) Cl + CCl3 → CCl4 (fast) What rate law does this mechanism predict?A)kG) [CHCl3]1/2M) [CCl3]2B) [Cl2]H) [CCl3]1/2N) [HCl]2C) [Cl]I) [HCl]1/2O) [Cl2]2D) [CHCl3]J) [Cl2]1/2P) [Cl]2E) [CCl3]K) [Cl]1/2 F) [HCl]L) [CHCl3]2

The following mechanism has been proposed for the gas-phase reaction of chloroform (CHCl3) and chlorine.Cl2 ⇌ 2Cl (fast, reversible) Cl + CHCl3 → HCl + CCl3 (slow) Cl + CCl3 → CCl4 (fast) What rate law does this mechanism predict?A)kG) [CHCl3]1/2M) [CCl3]2B) [Cl2]H) [CCl3]1/2N) [HCl]2C) [Cl]I) [HCl]1/2O) [Cl2]2D) [CHCl3]J) [Cl2]1/2P) [Cl]2E) [CCl3]K) [Cl]1/2 F) [HCl]L) [CHCl3]2

Question

Asked Oct 30, 2019

2468 views

The following mechanism has been proposed for the gas-phase reaction of chloroform (CHCl3) and chlorine.

Cl2 ⇌ 2Cl (fast, reversible)

Cl + CHCl3 → HCl + CCl3 (slow)

Cl + CCl3 → CCl4 (fast)

What rate law does this mechanism predict?

A)k G) [CHCl3]1/2 M) [CCl3]2

B) [Cl2] H) [CCl3]1/2 N) [HCl]2

C) [Cl] I) [HCl]1/2 O) [Cl2]2

D) [CHCl3] J) [Cl2]1/2 P) [Cl]2

E) [CCl3] K) [Cl]1/2

F) [HCl] L) [CHCl3]2

check_circle

Expert Answer

Step 1

The three step mechanism of the given reaction is,

CL2 dissociate to give chorine a fast and reverse reaction.

React with CHCL3 to give HCL and CCL3 a slow reaction

React with CL to give CCL4.

Step 2

Add equation (1), (2), and (3) to get the overall reaction.

The overall reaction is expressed as,

CL2 dissociate to give chorine a fast and reverse reaction.

React with CHCL3 to give HCL and CCL3 a slow reaction

React with CL to give CCL4.

Step 3

The chlorine atom produced in equation (1) is consumed in equation (3) and CCl3 a molecule is produced in equation (2) is consumed in equation (3).

CL react with CHCL3 to give a reversible reaction of HCL and CCL4.

6 0
3 years ago
Starting with a 6.847 M stock solution of HNO3, five standard solutions are prepared via serial dilution. At each stage, 25.00 m
stiks02 [169]

Answer: The concentration of HNO_3 in the final solution is 0.006688 M and number of moles are 0.00006688

Explanation:

According to the neutralization law,

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 6.847 M

V_1 = volume of stock  solution = 25.00 ml

M_2 = molarity of ist dilute solution = ?

V_2 = volume of first dilute solution = 100.0 ml

6.847\times 25.00=M_2\times 100.0

M_2=1.712M

2) on second dilution;

1.712\times 25.00=M_2\times 100.0

M_2=0.4280M

3) on third dilution

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M_2=0.1070M

4) on fourth dilution

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M_2=0.02675M

5) on fifth dilution

0.02675\times 25.00=M_2\times 100.0

M_2=0.006688M

Thus the concentration of HNO_3 in the final solution is 0.006688 M

moles of HNO_3 = Molarity\times {\text {Volume in L}}=0.006688\times 0.01L=0.00006688moles

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4 years ago
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Vinvika [58]
The correct answer is option 1. Butane and 2-butene have the same total number of carbon atoms. They both have four carbon atoms. They differ in there structure since the latter has double bonds on it. As a result of the different structure, they also have different properties.
8 0
3 years ago
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