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miv72 [106K]
3 years ago
12

25) Which of the following would exert the most pressure on the ground? 3 points

Chemistry
2 answers:
Zarrin [17]3 years ago
8 0

Answer:

a women standing in high heels

Explanation:

ZanzabumX [31]3 years ago
4 0

Answer:

A woman standing in high heels

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First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta
Klio2033 [76]

Rate equation for first order reaction is as follows:

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}

Here, k is rate constant of the reaction, t is time of the reaction, A_{0} is initial concentration and A_{t} is concentration at time t.

The rate constant of the reaction is 0.1 day^{-1}.

(a) Let the initial concentration be 100, If 90% of the chemical is destroyed, the chemical present at time t will be 100-90=10, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days

Thus, time required to destroy 90% of the chemical is 23.03 days.

(b) Let the initial concentration be 100, If 99% of the chemical is destroyed, the chemical present at time t will be 100-99=1, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days

Thus, time required to destroy 99% of the chemical is 46.06 days.

(c)  Let the initial concentration be 100, If 99.9% of the chemical is destroyed, the chemical present at time t will be 100-99.9=0.1, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days

Thus, time required to destroy 99.9% of the chemical is 69.09 days.

5 0
3 years ago
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