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Finger [1]
3 years ago
6

Consider the reaction represented by the equation?2SO2(g) + O2(g) 2SO3(g).?For the system at chemical equilibrium, which of the

following explains what happens after the volume of the reaction mixture is increased (assume constant temperature)?
a. The amount of SO3(g) increases and the value for K increases.
b. The amount of SO3(g) decreases and the value for K increases.
c. The amount of SO3(g) stays the same and the value for K decreases.
d. The amount of SO3(g) decreases and the value for K stays the same.
e. The amount of SO3(g) increases and the value for K stays the same
Chemistry
1 answer:
rewona [7]3 years ago
3 0

Answer:

b. The amount of SO3(g) decreases and the value for K increases.

Explanation:

Hello,

In this case, for the given reaction:

2SO_2(g) + O_2(g)\rightleftharpoons  2SO_3(g)

The change in the stoichiometric coefficient is:

\Delta \nu=2-2-1=-1

In such a way, since the reagents have more moles than the products, based on Le Chatelier's principle, if the volume is increased, the side with more moles is favored. In addition, since the formation of reagent is favored, K is diminished based on the law of mass action shown below:

K=\frac{[SO_3]_{eq}^2}{[SO_2]_{eq}^2[O_2]_{eq}}

Therefore the answer is:

b. The amount of SO3(g) decreases and the value for K increases.

Best regards.

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2 years ago
Please help me
Wittaler [7]

Answer:

pH = 6.999

The solution is acidic.

Explanation:

HBr is a strong acid, a very strong one.

In water, this acid is totally dissociated.

HBr + H₂O  →  H₃O⁺  +  Br⁻

We can think pH, as - log 7.75×10⁻¹² but this is 11.1

acid pH can't never be higher than 7.

We apply the charge balance:

[H⁺] = [Br⁻] + [OH⁻]

All the protons come from the bromide and the OH⁻ that come from water.

We can also think [OH⁻] = Kw / [H⁺] so:

[H⁺] = [Br⁻] + Kw / [H⁺]

Now, our unknown is [H⁺]

[H⁺] =  7.75×10⁻¹² + 1×10⁻¹⁴ / [H⁺]

[H⁺] = (7.75×10⁻¹² [H⁺] + 1×10⁻¹⁴) /  [H⁺]

This is quadratic equation:  [H⁺]² - 7.75×10⁻¹² [H⁺] - 1×10⁻¹⁴

a = 1 ; b = - 7.75×10⁻¹² ; c = -1×10⁻¹⁴

(-b +- √(b² - 4ac) / (2a)

[H⁺] = 1.000038751×10⁻⁷

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A very strong acid as HBr, in this case, it is so diluted that its pH is almost neutral.

8 0
3 years ago
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Lyrx [107]
What question there’s no question on here
5 0
1 year ago
A _____ solution contains more dissolved solute than a saturated solution at the same temperature.
klemol [59]
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5 0
3 years ago
If you have a solution that contains 46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH, what is the mole fraction of
irakobra [83]

Answer:

0.22

Explanation:

Given, Mass of C_{18}H_{21}NO_3 = 46.85 g

Molar mass of C_{18}H_{21}NO_3 = 299.4 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{46.85\ g}{299.4\ g/mol}

Moles\ of\ C_{18}H_{21}NO_3= 0.1565\ mol

Given, Mass of C_{2}H_{5}OH = 125.5 g

Molar mass of C_{2}H_{5}OH = 46.07 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{125.5\ g}{46.07\ g/mol}

Moles\ of\ C_{2}H_{5}OH= 0.5535\ mol

So, according to definition of mole fraction:

Mole\ fraction\ of\ codeine=\frac {n_{codeine}}{n_{codeine}+n_{ethanol}}

Mole\ fraction\ of\ codeine=\frac{0.1565}{0.1565+0.5535}=0.22

4 0
3 years ago
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