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Blizzard [7]
2 years ago
5

Please answer quickly!!

Chemistry
1 answer:
FromTheMoon [43]2 years ago
5 0

Answer:

i do not no but i do no .nb dncvejx euvcir

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What is the pH of a 0.028M solution (pH=-log(M))<br><br> A) 3.56<br> B)2.88<br> C)1.55<br> D)1
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A sample of milk kept at 25 °C is found to sour 40 times as rapidly as when it is kept at 4 °C. Estimate the activation energy f
irga5000 [103]

Answer:

120.575 kJ is the activation energy for the souring process.

Explanation:

The formula for an activation energy is given as:

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 25^oC = 40k

K_2 = rate constant at 4^oC = k

Ea = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 25^oC=273+25=298 K

T_2 = final temperature = 4^oC=273+4=277 K

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

\log (\frac{k}{40k})=\frac{Ea}{2.303\times 8.314 J/mol K}[\frac{1}{298K}-\frac{1}{277 K}]

E_a=120,575.61J=120.575 kJ

120.575 kJ is the activation energy for the souring process.

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