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jok3333 [9.3K]
2 years ago
15

The second order reaction A → Products takes 13.5 s for the concentration of A to decrease from 0.740 M to 0.319 M. What is the

value of k for this reaction?
Chemistry
1 answer:
Ivanshal [37]2 years ago
8 0

The rate constant of the second order reaction is 0.137 M-1s-1.

<h3>What is the rate constant?</h3>

For the second order reaction we can write;

1/[A] = kt + 1/[A]o

[A]o = initial concentration

[A] = final concentration

k = rate constant

t = time

Now;

1/0.319 = 13.5k + 1/ 0.740

1/0.319 - 1/0.740 = 13.5k

3.13 - 1.35 = 13k

k = 3.13 - 1.35/13

k = 0.137 M-1s-1

Learn more about second order reaction:brainly.com/question/12446045

#SPJ1

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3 years ago
sample of atmospheric gas collected at an industrial site is stored in a 250 mL amber glass bottle that has a pressure of 1.02 a
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Answer:- New pressure is 0.942 atm.

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Pressure is directly proportional to the kelvin temperature. The equation used here is:

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T_1 = 20.3 + 273.15 = 293.45 K

T_2 = -2.0 + 273.15 = 271.15 K

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Let's plug in the values in the equation and solve it for final pressure.

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So, the new pressure of the jar is 0.942 atm.


5 0
3 years ago
Determine the amount of heat(in Joules) needed to boil 5.25 grams of ice. (Assume standard conditions - the ice exists at zero d
Yuki888 [10]
Following are important constant that used in present calculations
Heat of fusion of H2O = 334 J/g 
<span>Heat of vaporization of H2O = 2257 J/g </span>
<span>Heat capacity of H2O = 4.18 J/gK 
</span>
Now, energy required for melting of ICE = <span>  334 X 5.25 = 1753.5 J .......(1)
Energy required for raising </span><span>the temperature water from 0 oC to 100 oC =  4.18 X 5.25 X 100 = 2195.18 J .............. (2)
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</span><span>
Thus, total heat energy required for entire process = (1) + (2)  + (3)
                                                                        = 1753.5 + 2195.18 + 11849.25
                                                                        = </span><span>15797.93 J 
</span><span>                                                                        = 15.8 kJ
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