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Savatey [412]
3 years ago
14

11) What is the volume of 4.92 grams of hydrogen gas at STP? (Show the work using dimensional analysis)

Chemistry
1 answer:
lorasvet [3.4K]3 years ago
4 0

Answer:

The volume of 4.92 grams of hydrogen gas at STP is 9.105 litter.

Explanation:

We know that, the molecular weight of hydrogen gas (H_{2} ) = 1\times 2 = 2.

Moles of hydrogen gas = \frac{weight}{molecular weight}  = \frac{4.92}{2} = 2.46 moles.

At STP, 1 mole of hydrogen gas = occupies 22.4 litter

    So, 2.46 mole of hydrogen gas = occupies \frac{22.4}{2.46} = 9.105 litter.

Hence, the volume of 2.46 grams of hydrogen gas at STP is 9.105 litter.

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Nonpolar compound would be symmetrical
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blagie [28]

1, When temperature is increased the volume will also increase. this is because the particles will gain kinetic energy and bombard the walls of the container of the gas at a higher frequency, therefore, for the pressure to remain constant as per Charles' law, the volume will have to increase so that the rate of bombardment remains constant. This is explained by the Charles law which states that the volume of a gas is directly proportional to the absolute temperature provided pressure remains constant.

2. When temperature is Decreased the volume will also Decrease. this is because the particles will loose kinetic energy and bombard the walls of the container of the gas less frequently, therefore, for the pressure to remain constant as per Charles' law, the volume will have to reduce so that the rate of bombardment remains constant. This is explained by the Charles law which states that the volume of a gas is directly proportional to the absolute temperature provided pressure remains constant.

3. When temperature is increased the pressure will increase. This is because the gas particles gain kinetic energy and bombard the walls of the container more frequently. this is according to Pressure law which states that for a constant volume of a gas the pressure is directly proportional to absolute temperature

4. When temperature is decreased, pressure will decrease, This is because the gas particles lose kinetic energy and bombard the walls of the container less frequently. this is according to Pressure law which states that for a constant volume of a gas the pressure is directly proportional to absolute temperature

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3 years ago
Explain why are all valence electrons are either in an s or p orbital.
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4 0
3 years ago
A 17.11 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 21.71 g CO2 and 5.
Andru [333]

Answer:  The empirical formula and the molecular formula of the organic compound is CHO and C_4H_4O_4 respectively.

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 = 21.71 g

Mass of H_2O= 5.926 g

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 21.71 g of carbon dioxide, =\frac{12}{44}\times 21.71=5.921g of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 5.926 g of water, =\frac{2}{18}\times 5.926=0.658g of hydrogen will be contained.

Mass of oxygen in the compound = (17.11) - (5.921+0.658) = 10.53  g

Mass of C = 5.921 g

Mass of H = 0.658 g

Mass of O = 10.53 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.921g}{12g/mole}=0.493moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.658g}{1g/mole}=0.658moles

Mass of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{10.53g}{16g/mole}=0.658moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.493}{0.493}=1

For H =\frac{0.658}{0.493}=1

For O=\frac{0.658}{0.493}=1

The ratio of C : H: O =  1: 1: 1

Hence the empirical formula is CHO.

empirical mass of CHO = 12(1) + 1(1) + 1 (16) = 29

Molecular mass = 104.1 g/mol

n=\frac{\text {Molecular mass}}{\text {Equivalent mass}}=\frac{104.1}{29}=4

Thus molecular formula = n\times {\text {Empirical formula}}=4\times CHO=C_4H_4O_4

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