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Karo-lina-s [1.5K]
3 years ago
15

Suppose you are investigating how the amount of exercise a hamster gets affects how long the hamster lives. In order to control

the experiment, which of these should you do?
A. Give all the hamsters the exact same diet.
B. Compare how long hamsters live to how long mice live.
C. Have your friend do the same experiment with hamsters that you are doing.
D. Make all the hamsters run on a hamster wheel for the same amount of time.
Chemistry
2 answers:
ozzi3 years ago
4 0
Hello, to answer your question fully.

<span>D. Make all the hamsters run on a hamster wheel for the same amount of time.

 Because your doing the experiment on exercise not anything else 
 so keeping charts in a notebook and set a time on how long they were on it and how long you want them to be on it. 
   
                Signed by, Virtuoso Sargedog</span>

jekas [21]3 years ago
3 0

Answer:

D. Make all the hamsters run on a hamster wheel for the same amount of time.

Explanation:

Hello,

In this case, for you to study how the exercise affects the hamsters lifetime, you must put all the hamsters under the same exercising conditions, thus, by making all the hamsters run on a hamster wheel for the same amount of time, you can attain it.

Best regards.

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The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t
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The expression used with catalyst and without catalyst is,

\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}

\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}

where,

K_2 = rate of reaction with catalyst

K_1 = rate of reaction without catalyst

Ea_2 = activation energy with catalyst  = 59.0 kJ/mol = 59000 J/mol

Ea_1 = activation energy without catalyst  = 184 kJ/mol = 184000 J/mol

R = gas constant

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Now put all the given values in this formula, we get:

\frac{K_2}{K_1}=e^{\frac{184,000 kJ-59000 kJ}{R\times 300}}=7.632\times 10^{10}

The reaction enhances by 7.632\times 10^{10}  when catalyst is present.

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x=\frac{3900 year}{7.632\times 10^{10}}=5.11\times 10^{-8} years

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