How much work would it take to push two protons very slowly from a separation of 2.00×10−10m (a typical atomic distance) to 3.00
1 answer:
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<u>Answer:</u> The heat released for the given process is -1892 kJ
<u>Explanation:</u>
The processes involved in the given problem are:
Pressure is taken as constant.
To calculate the amount of heat released at same temperature, we use the equation:
......(1)
where,
q = amount of heat released = ?
m = mass of water/ice
= latent heat of fusion or vaporization
To calculate the amount of heat released at different temperature, we use the equation:
.......(1)
where,
q = amount of heat released = ?
= specific heat capacity of medium
m = mass of water/ice
= final temperature
= initial temperature
Calculating the heat absorbed for each process:
- <u>For process 1:</u>
Converting the latent heat of fusion in J/kg, we use the conversion factor:
1 kJ = 1000 J
So,
We are given:
Putting values in equation 1, we get:
- <u>For process 2:</u>
We are given:
Putting values in equation 2, we get:
Total heat absorbed =
Total heat absorbed =
Hence, the heat released for the given process is -1892 kJ
The concentration of A will be <em>0.34 mol·L⁻¹</em> after 60 min.
In a first-order reaction, the formula for the amount remaining after <em>n</em> half-lives is
If
∴