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Otrada [13]
3 years ago
9

What is the density of a 12.5-gram object that displaces 16 cubed centimeters of water?

Chemistry
1 answer:
lord [1]3 years ago
8 0
Density = mass/volume
Density =12.5/16=0.78125 g/cm^3

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When heat is applied to 80 grams of CaCO3, it yields 39 grams of B. Determine the percentage of the yield.
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The question is incomplete, the complete question is:

When heat is applied to 80 grams of CaCO3, it yields 39 grams of CaO Determine the percentage of the yield.

CaCO3→CaO + CO2

<u>Answer:</u> The % yield of the product is 87.05 %

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

We are given:

Given mass of CaCO_3 = 80 g

Molar mass of CaCO_3 = 100 g/mol

Putting values in equation 1, we get:

\text{Moles of }CaCO_3=\frac{80g}{100g/mol}=0.8mol

For the given chemical reaction:

CaCO_3\rightarrow CaO+CO_2

By stoichiometry of the reaction:

If 1 mole of CaCO_3 produces 1 mole of CaO

So, 0.8 moles of CaCO_3 will produce = \frac{1}{1}\times 0.8=0.8mol of CaO

We know, molar mass of CaO = 56 g/mol

Putting values in above equation, we get:

\text{Mass of CaO}=(0.8mol\times 56g/mol)=44.8g

The percent yield of a reaction is calculated by using an equation:

\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100 ......(2)

Given values:

Actual value of the product = 39 g

Theoretical value of the product = 44.8 g

Plugging values in equation 2:

\% \text{yield}=\frac{39 g}{44.8g}\times 100\\\\\% \text{yield}=87.05\%

Hence, the % yield of the product is 87.05 %

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