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IRISSAK [1]
3 years ago
6

A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C. Br2(g) + I2(g) 2IBr

(g) When the mixture has come to equilibrium, the concentration of iodine monobromide is 1.190 M. What is the equilibrium constant for this reaction at 350°C?
Chemistry
1 answer:
blagie [28]3 years ago
5 0

Answer:

The answer to your question is K = 1.48

Explanation:

Balanced chemical reaction

                   Br₂   +   I₂    ⇒    2IBr

Process

1.- Calculate the concentration of each of the reactants

[Br₂] = 0.6 mol  / 1  L

        = 0.6 M

[I₂] = 1.6 mol / 1 L

     = 1.6 M

[IBr] = 1.19 M

2.- Write the equation of the Equilibrium constant

      K = [IBr]² / [Br₂][I₂]

-Substitution

      K = [1.19]² / [0.6][1.6]

- Simplification

      K = 1.4161 / 0.96

- Result

      K = 1.48

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LekaFEV [45]

Answer:

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Explanation:

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If we require 2Al2O3

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8 0
2 years ago
Sodium hydrogen carbonate NaHCO3, also known as sodium bicarbonate or "baking soda", can be used to relieve acid indigestion. Ac
olga2289 [7]

Answer:

1.4952 grams of sodium bicarbonate she  would need to ingest to neutralize this much HCl.

Explanation:

Moles (n)=Molarity(M)\times Volume (L)

Moles of hydrochloric acid = n

Volume of hydrochloric acid solution = 200.0 mL = 0.200 L

Molarity of the hydrochloric acid = 0.089 M

n=0.089 M\times 0.200 L=0.0178 mol of HCL

HCl(aq)+NaHCO_3(aq)\rightarrow NaCl(aq)+H_2O(l)+CO_2(g)

According to reaction, 1 mole of HCl is neutralized by 1 mole of sodium bicarbonate.

Then 0.0178 moles of HCl wil be neutralized by :

\frac{1}{1}\times 0.0178 mol=0.0178 mol of sodium bicarbonate

Mass of 0.0178 moles of sodium bicarbonate:

0.0178 mol × 72 g/mol = 1.4952 g

1.4952 grams of sodium bicarbonate she  would need to ingest to neutralize this much HCl.

8 0
3 years ago
What is the molar mass of AgF. 51.9g/mol 198.8g/mol 126.87g/mol 99.7g/mol
lisov135 [29]
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3 0
3 years ago
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3 years ago
What volume of 24% trichloroacetic acid (tca) is needed to prepare eight 3 ounce bottles of 10% tca solution?
Sever21 [200]

Answer:

295.7 mL of 24% trichloroacetic acid (tca) is needed .

Explanation:

Let the volume of 24% trichloroacetic acid solution be x

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Amount of trichloroacetic acid in 24% solution of x volume of solution will be equal to amount of trichloroacetic acid in 10% solution of volume 709.68 mL.

x\times \frac{24}{100}=709.68 mL\times \frac{10}{100}

x = 295.7 mL

295.7 mL of 24% trichloroacetic acid (tca) is needed .

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3 years ago
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