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Georgia [21]
3 years ago
7

How many grams of O2 must be placed in a 3.33 liter container in order to exert a pressure of 2.46 atm at 27°C?

Chemistry
1 answer:
stepan [7]3 years ago
7 0

Answer:

10.656 g

Explanation:

Given data:

Mass of oxygen = ?

Volume of container = 3.33 L

Pressure of gas = 2.46 atm

Temperature = 27°C

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will convert the temperature.

27+273 = 300 K

Now we will put the values in formula.

n = PV/RT

n = 2.46 atm × 3.33 L /  0.0821 atm.L/ mol.K   × 300 K

n = 8.192 atm. L /24.63 atm.L/ mol

n = 0.333 mol

Mass in grams:

Mass = number of moles × molar mass

Mass = 0.333 mol × 32 g/mol

Mass = 10.656 g

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3 years ago
Select the correct answer
amm1812

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  C  The experiment shows that the red substance experienced a chemical change.

Explanation:

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3 years ago
a) Calculate the mass of Li formed by electrolysis of molten LiCl by a current of 8.2 104 A flowing for a period of 14 hr. Assum
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a) Answer:

= 210,000 gram or 210 Kg  

Explanation:

14 hr is equivalent to;

= 14 × 3600 s

But; 

8.2x10^4 A = 8.2x10^4 Coulomb /second = 8.2x10^4 C/s  

1 Faraday = 96485C  

Molar mass of Li: 6.941 g/mol  

Therefore;

{(8.2x10^4 C/s)×(14×3600s)×(6.941 g/mol)/(96485C/mol)}×70%  

= 2.1x10^5 gram  or 210 kg

b.  Answer;

= 0.23 kWh/mol

Explanation and solution;

Energy per mole is given by;

(96485C/mol) × (8.4V) = 8.10x10^5 (J/mol)  

= 8.10x10^5 (Ws/mol)  

Divide by 3600s/h to get:  

= 8.10x10^5 (Ws/mol)

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