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4 years ago

Please someone should explain gas laws to me

Chemistry

1 answer:

dimaraw [331]4 years ago

5 0

Answer:

Gas laws, Laws that relate the pressure, volume, and temperature of a gas. ... These two laws can be combined to form a single generalization of the behavior of gases known as an equation of state, PV = nRT, where n is the number of gram-moles of a gas and R is called the universal gas constant.

Explanation:

Hope it helps

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How do everyday activities like driving a car and using electricity lead to global warming

natima [27]

They produce a temporary surge of CO2 which is assimilated by flora and results in temp growth rate flux

4 0

4 years ago

A compound contains 1.2 g of carbon, 3.2 g of oxygen and 0.2g of hydrogen. Find the formula of the compound

Karolina [17]

Answer:

The empirical formula of the compound is $C_{0.504}HO_{1.008}$ .

Explanation:

We need to determine the empirical formula in its simplest form, where hydrogen ( $H$ ) is scaled up to a mole, since it has the molar mass, and both carbon ( $C$ ) and oxygen ( $O$ ) are also scaled up in the same magnitude. The empirical formula is of the form:

$C_{x}HO_{y}$

Where $x$ , $y$ are the number of moles of the carbon and oxygen, respectively.

The scale factor ( $r$ ), no unit, is calculated by the following formula:

$r = \frac{M_{H}}{m_{H}}$ (1)

Where:

$m_{H}$ - Mass of hydrogen, in grams.

$M_{H}$ - Molar mass of hydrogen, in grams per mole.

If we know that $M_{H} = 1.008\,\frac{g}{mol}$ and $m_{H} = 0.2\,g$ , then the scale factor is:

$r = \frac{1.008}{0.2}$

$r = 5.04$

The molar masses of carbon ( $M_{C}$ ) and oxygen ( $M_{O}$ ) are $12.011\,\frac{g}{mol}$ and $15.999\,\frac{g}{mol}$ , then, the respective numbers of moles are: ( $r = 5.04$ , $m_{C} = 1.2\,g$ , $m_{O} = 3.2\,g$ )

Carbon

$n_{C} = \frac{r\cdot m_{C}}{M_{C}}$ (2)

$n_{C} = \frac{(5.04)\cdot (1.2\,g)}{12.011\,\frac{g}{mol} }$

$n_{C} = 0.504\,moles$

Oxygen

$n_{O} = \frac{r\cdot m_{O}}{M_{O}}$ (3)

$n_{O} = \frac{(5.04)\cdot (3.2\,g)}{15.999\,\frac{g}{mol} }$

$n_{O} = 1.008\,moles$

Hence, the empirical formula of the compound is $C_{0.504}HO_{1.008}$ .

3 0

3 years ago

Calculate the percentage by mass of the indicated element in the following compounds.

jarptica [38.1K]

Answer: The mass percent of carbon in acetylene is 92.31 %.

Explanation:

In $C_2H_2$ , there are 2 carbon atoms and 2 hydrogen atoms.

To calculate the mass percent of element in a given compound, we use the formula:

$\text{Mass percent of carbon}=\frac{\text{Mass of carbon}}{\text{Molar mass of acetylene}}\times 100$

Mass of carbon = $2\times 12g/mol=24g$

Mass of acetylene = $2\times 12g/mol+2\times 1g/mol=26g$

Putting values in above equation, we get:-

$\text{Mass percent of carbon}=\frac{24}{26}\times 100=92.31\%$

Hence, the mass percent of carbon in acetylene is 92.31 %.

7 0

3 years ago

How many moles of hydrogen are in 4.70 moles of C12H22O11

Sphinxa [80]

Answer:

103.4 moles of hydrogen

Explanation:

Given data:

Moles of C₁₂H₂₂O₁₁ = 4.70 mol

Number of moles of hydrogen present = ?

Solution:

one mole of C₁₂H₂₂O₁₁ contain 22 moles of hydrogen. Thus in order to determine total number of moles in 4.70 moles of C₁₂H₂₂O₁₁ we will multiply the 22 moles of hydrogen with 4.70 moles of C₁₂H₂₂O₁₁.

1 mole of C₁₂H₂₂O₁₁ = 22 moles of Hydrogen

4.70 × 22 moles

103.4 moles of hydrogen

8 0

3 years ago

Show ur working:

AleksAgata [21]

Answer:

120 g SO3

Explanation:

128 g SO2 -> 160 g SO3

96 g SO2 -> x

x= (96 g SO2 * 160 g SO3)/128 g SO2

x= 120 g SO3

8 0

3 years ago