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slega [8]
3 years ago
10

Calculate the heat energy released when 24.1g of liquid Mercury at 25 degrees celsius is converted to solid? mercury.; Steve I g

ot that answer wrong. I have to convert grams to moles, find heat released in freezing and cooling and then add the amount of heat released Heat capacity for Hg(l)=28.0J/mol-K enthalpy of fusion: 2.29 kJ/mol
Chemistry
1 answer:
harkovskaia [24]3 years ago
7 0
M<span>olar mass Hg = 200.5920g/mol </span>
<span>Hg = 24.1/200.5920 = 0.120mol </span>

<span>234.32K liquid to 25C (298.15K) liquid </span>
<span>h = (0.120)(28.0)(298.15 - 234.32) = 214.4688 J </span>

<span>234.32K liquid to 234.32K solid </span>
<span>h = (0.120)(2.29) = 0.2748 kj </span>

<span>total heat = 0.2145 + 0.2748 = 0.4893 kJ</span>
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A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from
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Answer:

T_i~=163.1 ºC

Explanation:

We have to start with the variables of the problem:

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Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT

Q_A_u=m_A_u*Cp_A_u*deltaT

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Now we can <u>put the values into the equation</u>:

60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)

Now we can <u>solve for the initial temperature of gold</u>, so:

T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20

T_i~=163.1 ºC

I hope it helps!

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