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2 years ago

A material has a density of 8.9 g/cm^3. You have 3 grams of the substance. How much space does it take up?

1 answer:

4 0

Density (d) is the ratio of the substance mass (m) to its volume (v).
d = m / v
Rearranging the equation gives us the equation for volume which is,
v = m / d
Substituting the values,
v = 3 g / (8.9 g/cm³)
v = 0.34 cm³
The substance occupies approximately 0.34 cm³.

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Answer:

$210.7~g~MgCO_3$

Explanation:

We have to start with the reaction:

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If we take into account the number of atoms in the formula, we can calculate the molar mass of carbon dioxide:

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In other words: $1~mol~CO_2=~44~g~CO_2$ . With this in mind, we can calculate the moles:

$110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2$

Now, the molar ratio between carbon dioxide and magnesium carbonate is 1:1, so:

$2.5~mol~CO_2=2.5~mol~MgCO_3$

With the molar mass of $MgCO_3$ ( $(23.3*1)+(12*1)+(16*3)=84.3~g/mol$ . With this in mind, we can calculate the grams of magnesium carbonate:

$2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3$

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Answer:

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Explanation:

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