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prohojiy [21]
3 years ago
8

What is the molality of a solution made up of 43.6 mol of CACI₂ dissolved in 13.5 kg of water? Please Show work

Chemistry
1 answer:
Llana [10]3 years ago
6 0

Answer:

The answer to your question is m = 3.2

Explanation:

Molality is defined as the number of moles of a solute dissolved in a mass of solvent (kg).

Data

moles of solute = 43.6

mass of solvent = 13.5 kg

Formula

Molality = \frac{number of moles}{Kg of solvent}

Substitution

Molality = \frac{43.6}{13.5}

Simplification and result

Molality = 3.2

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Calculate ΔS° for the reaction: 4Cr(s) + 3O2(g) → 2Cr2O3(s), Substance: Cr(s) O2(g) Cr2O3(s), S°(J/K⋅mol): 23.77 205.138 81.2
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<u>Answer:</u> The value of \Delta S^o for the reaction is 1051.93 J/K

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\Delta S^o_{rxn}=[(2\times \Delta S^o_{(Cr_2O_3(s))})]-[(4\times \Delta S^o_{(Cr(s))})+(3\times \Delta S^o_{(O_2(g))})]

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\Delta S^o_{rxn}=[(2\times (881.2))]-[(4\times (23.77))+(3\times (205.13))]\\\\\Delta S^o_{rxn}=1051.93J/K

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