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3 years ago

For neon, determine the moles and mass contained in a 29.9−L volume at a pressure of 481.1 torr at a temperature of 300.0 K.

Chemistry

1 answer:

kondor19780726 [428]3 years ago

8 0

Answer:

.7689 mol

15.516 g

Explanation:

Use the Ideal Gas Law, PV = nRT.

Make sure to use the correct ideal gas constant R. You can either put R in torr, or you can change the pressure to atm. I've just used the torr ideal gas constant.

481.1 torr * 29.9 L = n 62.364 LTorr/molK * 300 K

14384.89 = 18709.2n

n = .7689 mol

The molar mass of neon (remember that neon gas = Ne, it's not diatomic) is 20.18 g/mol from the periodic table.

.7689 mol * 20.18 g/mol = 15.516 g

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Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib

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The given chemical equations follows:

Equation 1: $H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1$

Equation 2: $HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2$

The net equation follows:

$S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c$

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=\frac{1}{K_1}\times \frac{1}{K_2}$

We are given:

$K_1=9.57\times 10^{-8}$

$K_2=1.46\times 10^{-19}$

Putting values in above equation, we get:

$K_c=\frac{1}{(9.57\times 10^{-8})}\times \frac{1}{(1.46\times 10^{-19})}=7.16\times 10^{25}$

Hence, the value of $K_c$ for the final reaction is $7.16\times 10^{25}$

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