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2 years ago

The density of mercury is 13.6 g/cm3 . What volume (in quarts) is occupied by 100. g of Hg? (1 L = 1.06 qt)

Chemistry

1 answer:

SpyIntel [72]2 years ago

6 0

Answer:

0.00077 qt

Explanation:

Density -

Density of a substance is given by the mass of the substance divided by the volume of the substance .

Hence , d = m / V

V = volume

m = mass ,

d = density ,

From the question ,

The mass mercury = 100 g

Density of mercury = 13.6 g/cm³ .

Hence , by using the above formula ,and putting the corresponding values , the volume of mercury is calculated as -

d = m / V

13.6 g/cm³ = 100 g / V

V = 7.35 cm³

1 cm³ = 0.001 L

V = 7.35 * 0.001 L = 0.0073 L

Since ,

1 L = 1.06 qt

V = 0.0073* 1.06 qt = 0.0077 qt

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Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

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Mass of $H_2O=5.58g$

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In 44g of carbon dioxide, 12 g of carbon is contained.

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For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 5.58 g of water, $\frac{2}{18}\times 5.58=0.62g$ of hydrogen will be contained.

Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.206 moles.

For Carbon = $\frac{0.206}{0.206}=1$

For Hydrogen = $\frac{0.62}{0.206}=3$

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Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

Hence, the empirical formula for the given compound is $C_1H_{3}O_1=CH_3O$

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The equation used to calculate the valency is:

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

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$n=\frac{124g/mol}{31g/mol}=4$

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