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dusya [7]
3 years ago
10

How many moles of N20 gas would have a volume of 3.8 L at 460 mmHg and 77°C?

Chemistry
1 answer:
Citrus2011 [14]3 years ago
7 0

<u>We are given:</u>

Volume of gas = 3.8 L

Pressure = 460 mmHg

Temperature = 77°c = (77+273)K = 350K

<u>Converting the pressure to atm:</u>

Pressure(in atm)  = Pressure(in mmHg) / 760

Pressure = 460/760 = 0.6 atm

<u>Finding the number of moles:</u>

using the ideal gas equation:

PV = nRT                                         [where R is the universal gas constant]

<em>replacing the given values in this equation</em>

(0.6)(3.8) = n(0.082)(350)

n = (0.6*3.8)/(0.082*350)

n = 0.08 moles

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A toxicologist studying mustard gas, S(CH2CH2Cl)2, a blistering agent, prepares a mixture of 0.675 M SCl2and 0.973 M C2H4 and al
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Answer:

The value of K_p is 0.02495.

Explanation:

Initial concentration of SCL_2 gas = 0.675 M

Initial concentration of C_2H_4 gas = 0.973 M

Equilibrium concentration of mustard gas = 0.35 M

SCl_2 (g) + 2 C_2H_4 (g)\rightleftharpoons S(CH_2CH_2Cl)_2(g)

initially

0.675 M            0.973 M        0

At equilibrium ;

(0.675-0.35) M            (0.973-2 × 0.35) M        0.35 M

The equilibrium constant is given as :

K_c=\frac{[S(CH_2CH_2Cl)_2]}{[SCl_2][C_2H_4]^2}

=\frac{0.35 M}{(0.675-0.35) M\times ((0.973-2 × 0.35) M)^2}

K_c=14.45

The relation between K_p and K_c are :

K_p=K_c\times (RT)^{\Delta n}

where,

K_p = equilibrium constant at constant pressure = ?

K_c = equilibrium concentration constant =14.45

R = gas constant = 0.0821 L⋅atm/(K⋅mol)

T = temperature = 20.0°C =20.0 +273.15 K=293.15 K

\Delta n = change in the number of moles of gas = [(1) - (1 + 2)]=-2

Now put all the given values in the above relation, we get:

K_p=14.45\times (0.0821L.atm/K.mol\times 293.15 K)^{-2}

K_p=6.2\times 10^{4}

K_p=0.02495

The value of K_p is 0.02495.

7 0
3 years ago
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