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DaniilM [7]
3 years ago
5

What mass of propane is necessary to react with the amount of oxygen in the chemical formula C3H8 + 5O2 → 3CO2 + 4H2O

Chemistry
1 answer:
Aleks [24]3 years ago
6 0
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

n(O₂)=1.407 mol
M(C₃H₈)=44.1 g/mol

m(C₃H₈)/M(C₃H₈)=n(O₂)/5

m(C₃H₈)=M(C₃H₈)n(O₂)/5

m(C₃H₈)=44.1*1.407/5=12.410 g

the mass of propane is 12.410 g
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explain the relationship between the rate of effusion of a gas and its molar mass. methane gas (ch4) effuses 3.4 times faster th
Musya8 [376]

The molar mass of the unknown gas is 184.96 g/mol

<h3>Graham's law of diffusion </h3>

This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e

R ∝ 1/ √M

R₁/R₂ = √(M₂/M₁)

<h3>How to determine the molar mass of the unknown gas </h3>

The following data were obtained from the question:

  • Rate of unknown gas (R₁) = R
  • Rate of CH₄ (R₂) = 3.4R
  • Molar mass of CH₄ (M₂) = 16 g/mol
  • Molar mass of unknown gas (M₁) =?

The molar mass of the unknown gas can be obtained as follow:

R₁/R₂ = √(M₂/M₁)

R / 3.4R = √(16 / M₁)

1 / 3.4 = √(16 / M₁)

Square both side

(1 / 3.4)² = 16 / M₁

Cross multiply

(1 / 3.4)² × M₁ = 16

Divide both side by (1 / 3.4)²

M₁ = 16 / (1 / 3.4)²

M₁ = 184.96 g/mol

Learn more about Graham's law of diffusion:

brainly.com/question/14004529

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3 0
1 year ago
A mixture of water and acetone at 756 mm boils at 70.0°C. The vapor pressure of acetone is
Tatiana [17]

Given :

A mixture of water and acetone at 756 mm boils at 70.0°C.

The vapor pressure of acetone is  1.54 atm at 70.0°C, while the vapor pressure of water is 0.312 atm at the same temperature.

To Find :

The percentage composition of the mixture.

Solution :

By Raoult's law :

P=x_{acetone}P^o_{acetone}+x_{water}P^o_{water}\\\\x_{acetone}1.58+x_{water}0.312=\dfrac{756}{760}=0.995\ atm\\\\1.58x_a+0.312x_w=0.995......( 1 )

Also , x_a+x_b=1      ......( 2 )

Solving equation 1 and 2 , we get :

x_a=0.54\ and \ x_w=0.46 .

Mass of acetone ,

m_a=x_a\times MM_a\\\\m_a=0.54\times 58\\\\m_a=31.32\ g

Mass of water ,

m_w=x_w\times MM_w\\\\m_w=0.46\times 18\\\\m_a=8.28\ g

\%water =\dfrac{8.28}{8.28+31.32}\times 100\\\\\%water =20.9\%\\\\\%acetone =79.1\%

Hence , this is the required solution.

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3 years ago
. If sulphur (IV) oxide and methane are
zubka84 [21]

Answer:

The answer is C. 1:2

Explanation:

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