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3 years ago

If a sample of NH3 gas at 689,1 mmHg has a volume of 607 2 mL and the presure is

Chemistry

1 answer:

never [62]3 years ago

8 0

Answer:

The answer to your question is V2 = 746.1 ml

Explanation:

Data

Pressure 1 = P1 = 689.1 mmHg

Volume 1 = V1 = 607.2 ml

Pressure 2 = P2 = 560.8 mmHg

Volume 2 = V2 = ?

Process

To solve this problem use Boyle's law

P1V1 = P2V2

-Solve for V2

V2 = P1V1/P2

-Substitution

V2 = (689.1 x 607.2) / 560.8

-Simplification

V2 = 746.1 ml

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Why can't the subscripts be changed?

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Subscripts cannot be changed because they are the ratio of the amount. And as we know, in a chemical equation/reaction, mass cannot be created nor destroyed. Therefore, we cannot change subscripts, however, we could change coefficients.

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The negatively charged particles of an atom are the

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Answer:

Electron: A negatively charged particle found circling or orbiting an atomic nucleus. An electron, like a proton is a charged particle, although opposite in sign, but unlike a proton, an electron has negligible atomic mass. Electrons contribute no atomic mass units to the total atomic weight of an atom.

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3 years ago

Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is comb

Bogdan [553]

Answer: The molecular formula for the menthol is $C_{10}H_{20}O$

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:

Mass of $CO_2=0.2829g$

Mass of $H_2O=0.1159g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 0.2829 g of carbon dioxide, $\frac{12}{44}\times 0.2829=0.077g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.1159 g of water, $\frac{2}{18}\times 0.1159=0.0129g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1005) - (0.077 + 0.0129) = 0.0106 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0129g}{1g/mole}=0.0129moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0106g}{16g/mole}=0.00066moles$

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 0.00066 moles.

For Carbon = $\frac{0.0064}{0.00066}=9.69\approx 10$

For Hydrogen = $\frac{0.0129}{0.00064}=19.54\approx 20$

For Oxygen = $\frac{0.00066}{0.00066}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 10 : 20 : 1

Hence, the empirical formula for the given compound is $C_{10}H_{20}O_1=C_{10}H_{20}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

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

We are given:

Mass of molecular formula = 156.27 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

$n=\frac{156.27g/mol}{156g/mol}=1$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O$

Thus, the molecular formula for the menthol is $C_{10}H_{20}O$

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