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2 months ago

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An unsaturated aqueous solution of NH3 is at 90.°C in 100. grams of water. According to Reference Table G, how many grams of NH3

could this unsaturated solution contain?
(1) 5 g (3) 15 g
(2) 10. g (4) 20. g

1 answer:

Tems11 [2.7K]2 months ago

3 0

The result is 2. 10 g
I hope this helps you

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Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

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Answer: The molecular formula for the specified organic compound is $C_{18}H_{20}O_2$

Explanation:

The combustion reaction of a hydrocarbon comprising carbon, hydrogen, and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

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

The information provided includes:

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From our knowledge:

Molar mass of carbon dioxide is 44 g/mol

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For determining the amount of carbon:

In carbon dioxide weighing 44 g, 12 g of carbon is found.

Hence, in 39.61 g of carbon dioxide, $\frac{12}{44}\times 39.61=10.80g$ grams of carbon will be found.

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Thus, in 9.01 g of water, $\frac{2}{18}\times 9.01=1.00g$ grams of hydrogen will be present.

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To derive the empirical formula, the following steps must be followed:

Step 1: Convert the indicated masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles$

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

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles$

Step 2: Calculating the ratio of moles of the respective elements.

To find the mole ratio, each mole value is divided by the smallest amount of moles calculated, which is 0.10 moles.

For Carbon = $\frac{0.9}{0.10}=9$

For Hydrogen = $\frac{1}{0.10}=10$

For Oxygen = $\frac{0.10}{0.10}=1$

Step 3:Using the mole ratios as subscripts.

The ratio of C: H: O = 9: 10: 1

Therefore, the empirical formula for the mentioned compound is $C_9H_{10}O$

To ascertain the molecular formula, it is necessary to find the valency, which is multiplied by each element to derive the molecular formula.

The equation to determine the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

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$C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2$

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Best of luck

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