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3 years ago

11

25.2 mL of water are placed in a graduated cylinder. A 22.6 g stone is dropped in, and the water level rises to 32.4 mL. Find th

e stone's density.

1 answer:

VMariaS [17]3 years ago

5 0

<span>Displaced volume :

final volume - initial volume

32.4 mL - 25.2 mL => 7.2 mL

Density = mass / volume

D = 22.6 g / 7.2 mL

D = 3.1388 g/mL

hope this helps!</span>

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Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

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where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

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

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

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$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

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$\Delta S^o$ = entropy of reaction = ?

n = number of moles

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

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

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Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

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