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3 years ago

What concentration of so32– is in equilibrium with ag2so3(s) and 2.60 × 10-3 m ag ? the ksp of ag2so3 can be found here?

1 answer:

8 0

When the Ksp of Ag2SO3 is 1.5 x 10^-14

and when the reaction equation is:

Ag2SO3 ↔ 2Ag+(aq) + SO3-- (aq)

so,

Ksp = [Ag+]^2 *[SO3--]

by substitution:

1.5 10^-14 = (2.6 x 10^-3)^2 *[SO3--]

∴ [SO3--] = 2.2 x10^-9

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The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C

avanturin [10]

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Explanation :

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

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$T_{final}$ = final temperature = $27.43^oC$

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

$q=5.20kJ/^oC\times (27.43-22.93)^oC$

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$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

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$\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole$

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

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Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-2805.8kJ/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles = 0 (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = $27.43^oC=273+27.43=300.43K$

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

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K$

$\Delta U=-2805.8kJ/mol-0$

$\Delta U=-2805.8kJ/mol$

Therefore, the internal energy change is -2805.8 kJ/mol

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kolezko [41]

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Explanation:

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2 years ago

Read 2 more answers