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artcher [175]
2 years ago
12

Determine the percent water in MgSO4*7H20?

Chemistry
1 answer:
Semmy [17]2 years ago
5 0

Answer:

\% H_2O=51.2\%

Explanation:

Hello!

In this case, since the percent water is computed by dividing the amount of water by the total mass of the hydrate; we infer we first need the molar mass of water and that of the hydrate as shown below:

MM_{MgSO_4* 7H_2O}=120.36 g/mol+7*18.02g/mol\\\\MM_{MgSO_4* 7H_2O}=246.5g/mol

Thus, the percent water is:

\% H_2O=\frac{7*MM_{H_2O}}{MM_{MgSO_4* 7H_2O}} *100\%\\\\

So we plug in to obtain:

\% H_2O=\frac{7*18.02}{246.5} *100\%\\\\\% H_2O=51.2\%

Best regards!

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Read 2 more answers
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Alexxx [7]

91 grams of sodium azide required to decompose and produce 2.104 moles of nitrogen.

Explanation:

2NaN3======2Na+3N2

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From the equation:

2 moles of NaNO3 will undergo decomposition to produce 3 moles of nitrogen.

In the question moles of nitrogen produced is given as 2.104 moles

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From the stoichiometry,

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So, 1.4 moles of sodium azide will be required to decompose to produce 2.104 moles of nitrogen.

From the formula

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3 years ago
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2 years ago
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Answer:

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