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Alenkasestr [34]

3 years ago

10

Calculate the heat energy released when 13.3 g of liquid mercury at 25.00 C is converted to solid mercury at its melting point.C

onstants for mercury at 1 atmheat capacity of Hg(l) 28.0 J/(mol K)melting point 234.32 Kenthalphy of fusion 2.29 KJ/mol

1 answer:

dmitriy555 [2]3 years ago

4 0

Answer:

$-270.321012\ J$

Explanation:

$C_v$ = Heat capacity of Hg = 28 J/mol

$\Delta T$ = Change in temperature = $(234.32-(273.15+25))$

$\Delta H_{f}$ = Enthalpy of fusion = 2.29 kJ/mol

The number of moles is given by

$n=13.3\times \dfrac{1}{200.59}\\\Rightarrow n=0.0663\ molHg$

Heat is given by

$Q_1=nC_v\Delta T\\\Rightarrow Q_1=0.0663\times 28\times (234.32-(273.15+25))\\\Rightarrow Q_1=-118.494012\ J$

Heat released is given by

$Q_2=-n\Delta H_{f}\\\Rightarrow Q_2=-0.0663\times 2.29\times 10^3\\\Rightarrow Q_2=-151.827\ J$

Total heat is given by

$Q=Q_1+Q_2\\\Rightarrow Q=-118.494012+(-151.827)\\\Rightarrow Q=-270.321012\ J$

The total heat released is $-270.321012\ J$

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3 years ago

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Answer:

The displacement of the air drop after 3 second is 18.27 m.

Explanation:

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