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bazaltina [42]
3 years ago
14

This reaction was at equilibrium when 0.2 atm of iodine gas was pumped into the container, what happened to the equilibrium and

the partial pressures of the gases
Chemistry
1 answer:
n200080 [17]3 years ago
7 0

Answer:

Q was < K. Partial pressure of hydrogen decreased, iodine increased

Explanation:

<em>After iodine was added the Q was [Select] K so the reaction shifted toward the Products [Select] ,The partial pressure of hydrogen [Select], Iodine [Select] |,and hydrogen iodide Decreased</em>

Based on the equilibrium:

H2(g) + I2(g) ⇄ 2HI(g)

K of equilibrium is:

K = [HI]² / [H2] [I2]

<em>Where [] are concentrations at equilibrium</em>

And Q is:

Q = [HI]² / [H2] [I2]

<em>Where [] are actual concentrations of the reactants.</em>

<em />

When the reaction is in equilibrium, K=Q.

But as [I2] is increased, Q decreases and Q was < K

The only concentration that increases is [I2], doing partial pressure of hydrogen decreased, iodine increased

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Exactly how much time must elapse before 16 grams of potassium-42decays, leaving 2 grams of the original isotope?(1) 8 × 12.4 ho
AleksandrR [38]
The answer is <span>(3) 3 × 12.4 hours
</span>
To calculate this, we will use two equations:
(1/2) ^{n} =x
t_{1/2} = \frac{t}{n}
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>t_{1/2} - half-life length
</span>t - total time elapsed.

First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
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Thus:
<span>(1/2) ^{n} =x
</span>(0.5) ^{n} =0.125
n*log(0.5)=log(0.125)
n= \frac{log(0.5)}{log(0.125)}
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It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>t_{1/2} = 12.4
</span><span>t_{1/2} *n = t
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Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>
7 0
3 years ago
Read 2 more answers
it takes 513 kj to remove a mole of electrons from the atoms at the surface of a piece of metal. how much energy does it take to
schepotkina [342]

Answer:

The right solution is "8.5\times 10^{-19} \ joule".

Explanation:

As we know,

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= 8.5\times 10^{-19} \ joule

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