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Murrr4er [49]
2 years ago
9

What is the volume of 3.40 mol of gas at 33.3C and 22.2atm of pressure

Chemistry
1 answer:
Luda [366]2 years ago
5 0
The volume of 3.40 mol of gas at 33.3 C and 22.2 atm of pressure is 3.85 liter of gas. This problem can be solved by using the PV=nRT or V= nRT/V equation which is the relation between the molar volume, the temperature, and the pressure of gas. In this formula, P is the pressure, R is the universal gas constant ( 0.0821 atm L/mol K), n is the molecule amount, V is the molecular volume, and T is the temperature. The temperature used in this formula must be in Kelvin, therefore we have to convert the Celcius temperature into Kelvin temperature (33.3 C = 306.45 K).
<span>The calculation for the problem above: 3.4*0.0821*306.45/22.2 = 3.85 liter of gas.</span>
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Answer:

Both Option (I) and Option (III)

Explanation:

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3 years ago
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Answer:

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A closed system initially containing 1×10^-3 hydrogen 2×10^-3M iodine at 448 degree Celsius and is allowed to reach equilibrium.
GaryK [48]

Answer:

Kc = 50.5

Explanation:

We determine the reaction:

H₂  +  I₂   ⇄   2HI

Initially we have 0.001 molesof H₂

and 0.002 moles of I₂

If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.

           H₂     +      I₂      ⇄   2HI

In:     0.001       0.002           -

R:       x                 x                2x

Eq:  0.001-x    0.002-x      0.00187  

x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted

So in the equilibrium we have:

0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵  moles of H₂

0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂

Expression for Kc is =  (HI)² / (H₂) . (I₂)

0.00187 ² /  6.5×10⁻⁵ . 1.065×10⁻³ = 50.5

5 0
3 years ago
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