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ra1l [238]
3 years ago
9

What id the volume if a gas if I have 0.215 moles if gas at a temperature of 75 K and a pressure if 182.34KPa?

Chemistry
1 answer:
ratelena [41]3 years ago
7 0
This problem tests the concept of general gas equation, which is:

PV = nRT

Where,
P = Pressure
V = Volume
n = number of moles
R = General gas constant
T = Temprature

We are to find V, and we have the rest of the values.

P = 183.24 KPa = 183240 Pa
n = 0.215 moles
T = 75 K
R = 8.314 J /(mol.K)

Using the values in the equation, we get:

183240 V = 0.215*75*8.314

V = 0.000732 m³

The value of volume is cubic meter. In cubic cm, the value of volume will be:
 V = 0.000732 x 1000000 = 732 cm³
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A sealed can of your favorite soda has a carbon dioxide gas (C0) volume of 0.05 L. When it is refrigerated, the
liubo4ka [24]

Answer:

<h2>0.102 L</h2>

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question we have

V_2 =  \frac{207000 \times 0.05}{101000}  =  \frac{10350}{101000}  \\  = 0.102475...

We have the final answer as

<h3>0.102 L</h3>

Hope this helps you

8 0
3 years ago
How many moles of carbon are there in 5 g of carbon?
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Answer:

0.416666667

Explanation:

number of moles= mass of sample ÷ molar mass

=5÷12

=0.41666667

8 0
2 years ago
calculate the molality of commercial HCl solution which is 12.1Molarity, and has a density of 1.19g/mL, and is 37.2wt.%HCl.
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3 years ago
For the reaction 2 H2S(g) D 2 H2 (g) + S2 (g), Kp = 1.5 × 10−5 at 800.0°C. If the initial partial pressures of H2 and S2 in a cl
Usimov [2.4K]

Answer: The approximate equilibrium partial pressure of H_2S is 3.92  atm

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

      2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)

K_p=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}

1.5\times 10^{-5}=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}

On reversing the reaction:

     2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)

initial pressure  4.00atm    2.00 atm       0

eqm          (4.00-2x)atm      (2.00-x) atm      2x atm

K_p=\frac{[H_2S]^2}{[H_2]^2\times [S_2]}

K_p'=\frac{1}{K_p}=0.67\times 10^5

2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)

0.67\times 10^5=\frac{2x]^2}{[4.00-2x]^2\times [2.00-x]}

x=1.96

[H_2S]=2x=2\times 1.96=3.92 atm

Thus approximate equilibrium partial pressure of H_2S is 3.92 atm

3 0
3 years ago
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