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Contact [7]
3 years ago
8

4.3 moles of a gas are at a temperature of 28 degrees * C with a pressure of 1.631 atm. What volume does the gas occupy?

Chemistry
1 answer:
Shkiper50 [21]3 years ago
7 0

Answer:

65.2L

Explanation:

Using the general gas equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (Litres)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (Kelvin)

According to the information provided in this question,

P = 1.631 atm

V = ?

n = 4.3 moles

T = 28°C = 28 + 273 = 301K

Using PV = nRT

V = nRT/P

V = 4.3 × 0.0821 × 301 ÷ 1.631

V = 106.26 ÷ 1.631

V = 65.15

Volume of the gas = 65.2L

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Which properties change the composition of a substance?
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Answer:

A.

Explanation:

Physical properties are characteristics of a substance that can be observed or measured without changing the composition of the substance. Some examples of physical properties are its volume and its density. There are two types of physical properties, intensive physical properties and extensive physical properties.

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3 years ago
What causes the Brownian motion of particles in a colloid?
LekaFEV [45]

<u>Answer:</u>

The correct answer option is a) collisions between the particles and surrounding molecules.

<u>Explanation:</u>

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Brownian motion is the irregular movement of the microscopic particles in a fluid which bombard into each other.

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3 years ago
A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed
AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) <em>Balanced equation including N_2 from air</em>  

The balanced equation <em>ignoring</em> N_2 from air is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2  

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2  

<em>Including</em> N_2 from air, the balanced equation is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2  

(b) <em>Balanced equation for 120 % stoichiometric combustion</em>  

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2  

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2  

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2  

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) <em>Minimum mass of air</em>  

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2  

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2  

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2  

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air  

(d) <em>Air:fuel mass ratio for 100 % combustion</em>  

Air:fuel = 17 300 kg/1700 kg = <em>10.2 :1 </em>

(e) <em>Air:fuel mass ratio for 120 % combustion </em>

Mass of air = 17 300 kg × 1.20 = 20 760 kg air  

Air:fuel = 20 760 kg/1700 kg = 12.2 :1  

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