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!

5 0

2 years ago

Write an equation that shows the formation of a chromium(ii) ion from a neutral chromium atom.

daser333 [38]

The equation that shows the formation of chromium (ii) ion from neutral chromium atom is as follow
Cr ---> cr^2+ + 2e-
Cr^2+ is the chromium ion with oxidation state of two which is one of the common ion of chromium. Other common ion of chromium include chromium of oxidation state 6 and 3

8 0

3 years ago

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