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sveta [45]
3 years ago
8

Assume an object A mass of 36.240 g and it’s density is 5.40 g/mL. What is it’s volume

Chemistry
1 answer:
topjm [15]3 years ago
5 0

The volume of the object=6.71ml

Given:

Mass of the object=36.240g

Density of the object=5.40\frac{g}{m l}

To find:

Volume of the object

<u>Step by Step Explanation:</u>

Solution:

According to the formula, Density of the object is given as

\rho=\frac{m}{v} and from this volume can be derived as

v=\frac{m}{\rho}

Where, m=mass of the cylinder

\rho=density of the cylinder

v=volume of the cylinder

We know the values of m=36.240g and \rho=5.40\frac{g}{m l}

Substitute these values in the above equation we get

v=\frac{m}{\rho}

v=36.240/5.40

=6.71ml

Result:

Thus the volume of the object is 6.71ml

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Consider the balanced equation for the following reaction:
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<u>Answer:</u> The amount of carbon dioxide formed in the reaction is 5.663 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

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

Given mass of oxygen gas = 8 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{8g}{32g/mol}=0.25mol

For the given chemical equation:

7O_2(g)+2C_2H_6(g)\rightarrow 4CO_2(g)+6H_2O(l)

By Stoichiometry of the reaction:

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So, 0.25 moles of oxygen gas will produce = \frac{4}{7}\times 0.25=0.143mol of carbon dioxide

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Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.143 moles

Putting values in equation 1, we get:

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To calculate the experimental yield of carbon dioxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Percentage yield of carbon dioxide = 90 %

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Putting values in above equation, we get:

90=\frac{\text{Experimental yield of carbon dioxide}}{6.292g}\times 100\\\\\text{Experimental yield of carbon dioxide}=\frac{90\times 6.292}{100}=5.663g

Hence, the amount of carbon dioxide formed in the reaction is 5.663 grams

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