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const2013 [10]
3 years ago
6

For the equilibrium

Chemistry
1 answer:
Mamont248 [21]3 years ago
3 0

Answer:

Equilibrium concentrations of the gases are

H_2S=0.596M

H_2=0.004 M

S_2=0.002 M

Explanation:

We are given that  for the equilibrium

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

k_c=9.0\times 10^{-8}

Temperature, T=700^{\circ}C

Initial concentration of

H_2S=0.30M

H_2=0.30 M

S_2=0.150 M

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

H_2S=0.30+2x

H_2=0.30-2x

S_2=0.150-x

At equilibrium

Equilibrium constant

K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}

Substitute the values

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

By solving we get

x\approx 0.148

Now, equilibrium concentration  of gases

H_2S=0.30+2(0.148)=0.596M

H_2=0.30-2(0.148)=0.004 M

S_2=0.150-0.148=0.002 M

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Brilliant_brown [7]

Answer:

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Explanation:

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In this case, when focusing on these heating curves, it is important to say they tend to have two constant-temperature sections and three variable-temperature sections. Thus, from lower to higher temperature, the first constant-temperature section corresponds to melting and the second one vaporization, whereas the three variable-temperature sections correspond to the heating of the solid until melting, the liquid until vaporization and the gas until the critical point.

In such a way, we infer that the boxes referred to constant temperature are referred to a gain in potential energy, that is, the two horizontal lines.

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How many molecules of H2O are there in 1.0 g of H2O?
NISA [10]
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8 0
3 years ago
A 25.0 mL sample of 0.123 M HCl was titrated with 18.04 mL of NaOH . What is the concentration of the NaOH ?
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Find the pH. What are the pH values for the following solutions? (a) 0.1 M HCl (b) 0.1 M NaOH (c) 0.05 M HCl (d) 0.05 M NaOH
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Answer:

(a) pH=1

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(c) pH=13

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Explanation:

Hello,

In this case, we define the pH in terms of the concentration of hydronium ions as:

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Which is directly computed for the strong hydrochloric acid (consider a complete dissociation which means the concentration of hydronium equals the concentration of acid) in (a) and (c) as shown below:

(a)

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pH=-log(0.1)=1

(b)

[H^+]=[HCl]=0.05M

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Nevertheless, for the strong sodium hydroxide, we don't directly compute the pH but the pOH since the concentration of base equals the concentration hydroxyl in the solution:

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(b)

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(d)

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