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Lelechka [254]
2 years ago
6

A solution made by mixing 20.0 g of a non-volatile compound with 125 mL of water at 25°C has a vapor pressure of 22.67 torr. Wha

t is the molecular weight (3sf) of the compound (g/mol)? (Do not include units in your answer. JUST THE NUMBER
Chemistry
1 answer:
Ainat [17]2 years ago
4 0

We have that the molecular weight (3sf) of the compound (g/mol)

m=44.15g/mol

From the question we are told

A solution made by mixing 20.0 g of a non-volatile compound with 125 mL of water at 25°C has a vapor pressure of 22.67 torr. What is the molecular weight (3sf) of the compound (g/mol).

Generally the equation for the Rouault's law is mathematically given as

P=P_0 N

22.67=23.8*\frac{\frac{12.5}{18}}{\frac{125}{18}+\frac{15}{m}}\\\\\6.95+\frac{15}{m}=7.29\\\\\frac{15}{m}=7.29-6.95\\\\m=\frac{15}{0.34}\\\\m=44.11g/mol

Therefore

The molecular weight (3sf) of the compound (g/mol)

m=44.15g/mol

For more information on this visit

brainly.com/question/17756498

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Answer:

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Explanation:

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Mg²⁺  + 2e⁻ → Mg

Si  → SiO₃²⁻ + 4e⁻

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If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

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Now, that they are ballanced we can sum the half reactions:

2Mg²⁺  + 4e⁻ → 2Mg

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The oxidation number of cation, K is +1. Hence, the total charge of the anion, [Fe(CN)₆] is -4. CN has charge has -1. There are 6 CN in anion. Let's assume the oxidation number of Fe is 'a'.

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