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astraxan [27]
2 years ago
11

A 16 gram sample of O2(g) fills a container at STP. What volume is the container?

Chemistry
1 answer:
sergij07 [2.7K]2 years ago
8 0

Answer:

V = 11.2 L

Explanation:

Hello there!

In this case, according to the ideal gas equation:

PV=nRT

It is possible to compute the volume as shown below:

V=\frac{nRT}{P}

Whereas the moles are computed are computed given the mass and molar mass of oxygen:

n=16g*\frac{1mol}{32g} =0.5mol

Now, since the STP stands for a temperature of 273.15 K and a pressure of 1 atm, the resulting volume is:

V=\frac{0.5mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm}\\\\V=11.2L

Best regards!

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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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<u>Answer:</u> The value of K_c for the final reaction is 7.16\times 10^{25}

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u>  H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1

<u>Equation 2:</u>  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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