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Triss [41]
3 years ago
5

A 0.055 mol sample of formaldehyde vapor, CH2O, was placed in a heated 500 mL vessel and some of it decomposed. The reaction is

CH2O(g) H2(g) + CO(g) At equilibrium, the CH2O(g) concentration was 0.051 mol L-1. Calculate the value of Kc for this reaction?
Chemistry
1 answer:
Elis [28]3 years ago
7 0

Answer:

Kc for this reaction is 0.06825

Explanation:

Step 1: Data given

Number of moles formaldehyde CH2O = 0.055 moles

Volume = 500 mL = 0.500 L

At equilibrium, the CH2O(g) concentration = 0.051 mol

Step 2: The balanced equation

CH2O  <=>  H2 + CO

Step 3: Calculate the initial concentrations

Concentration = moles / volume

[CH2O] = 0.055 moles . 0.500 L

[CH2O] = 0.11 M

[H2] = 0M

[CO] = 0M

Step 4: The concentration at the equilibrium

[CH2O] = 0.11 - X M = 0.051 M

[H2] = XM

[CO] = XM

[CH2O] = 0.11 - X M = 0.051 M

X = 0.11 - 0.051 = 0.059

[H2] = XM = 0.059 M

[CO] = XM = 0.059 M

Step 5: Calculate Kc

Kc = [H2][CO]/[CHO]

Kc = (0.059 * 0.059) / 0.051

Kc = 0.06825

Kc for this reaction is 0.06825

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3 years ago
the rate of disappearance of Br- at some moment in time was determined to be 3.5 x 10-4 M/s. What is the rate of appearance of B
ddd [48]

Answer:

1.8 × 10⁻⁴ mol M/s

Explanation:

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Step 2: Establish the appropriate molar ratio

The molar ratio of Br⁻ to Br₂ is 2:1.

Step 3: Calculate the rate of appearance of Br₂

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3 0
3 years ago
Atoms of which of the following elements are largest?<br> A:<br> Na<br> OOOO
elena-14-01-66 [18.8K]

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5 0
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Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit
Rasek [7]

Answer:

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Explanation:

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0.0811 M=\frac{n}{0.0017 L}

n = 0.0001378 mol

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According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0
3 years ago
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