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3 years ago

How many milliliters of an aqueous solution of 0.222 M manganese(II) iodide is needed to obtain 9.57 grams of the salt?

Chemistry

1 answer:

Vaselesa [24]3 years ago

8 0

Answer:

139 mL

Explanation:

Given data:

Volume of solution required = ?

Molarity of solution = 0.222 M

Mass of salt = 9.57 g

Solution:

Number of moles of manganese iodide:

Number of moles = mass/molar mass

Number of moles = 9.57 g/ 308.75 g/mol

Number of moles = 0.031 mol

Volume required:

Molarity = number of moles of solute / volume in L

0.222 M = 0.031 mol / V

V = 0.031 mol / 0.222 mol/L

V = 0.139 L

Volume in mL:

0.139 L ×1000 mL / 1 L

139 mL

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2 years ago

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Explanation:

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The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]$

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Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ$

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9. If the concentration of H2O (a reactant) is decreased and that of CO (a product) is increased, both actions lead to the equilibrium being shifted to the left.

10. Addition of catalyst to the system will only speed up the rate at which the system reach the equilibrium.

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2 years ago

How many milliliters of an aqueous solution of 0.222 M manganese(II) iodide is needed to obtain 9.57 grams of the salt?​

How many milliliters of an aqueous solution of 0.222 M manganese(II) iodide is needed to obtain 9.57 grams of the salt?