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Over [174]
3 years ago
10

The gases carbon dioxide, oxygen, and argon are mixed in a container. All gases have the same partial pressure, and the total pr

essure of the container is 32,680 Pa. What is the partial pressure of argon ?
Chemistry
1 answer:
polet [3.4K]3 years ago
4 0
The partial pressure of argon can be calculated simply by dividing the total pressure by 3. This gives an answer equivalent to 10893.33 Pa. The clue behind this is the statement "a<span>ll gases have the same partial pressure". This follows Dalton's Law of Partial Pressures which states that the total pressure is equal to the sum of the individual partial pressures.</span>
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C2H6(g) + O2(g) → CO2(g) + H2O(g)
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Be sure to answer all parts. A 0.365−mol sample of HX is dissolved in enough H2O to form 835.0 mL of solution. If the pH of the
Marta_Voda [28]

Answer:

The Ka is 9.11 *10^-8

Explanation:

<u>Step 1: </u>Data given

Moles of HX = 0.365

Volume of the solution = 835.0 mL = 0.835 L

pH of the solution = 3.70

<u>Step 2:</u> Calculate molarity of HX

Molarity HX = moles HX / volume solution

Molarity HX = 0.365 mol / 0.835 L

Molarity HX = 0.437 M

<u />

<u>Step 3:</u> ICE-chart

[H+] = [H3O+] = 10^-3.70 = 1.995 *10^-4

Initial concentration of HX = 0.437 M

Initial concentration of X- and H3O+ = 0M

Since the mole ratio is 1:1; there will react x M

The concentration at the equilibrium is:

[HX] = (0.437 - x)M

[X-] = x M

[H3O+] = 1.995*10^-4 M

Since 0+x = 1.995*10^-4   ⇒ x=1.995*10^-4

[HX] = 0.437 - 1.995*10^-4 ≈ 0.437 M

[X-] = x = 1.995*10^-4 M

<u>Step 4: </u>Calculate Ka

Ka = [X-]*[H3O+] / [HX]

Ka = ((1.995*10^-4)²)/ 0.437

Ka = 9.11 *10^-8

The Ka is 9.11 *10^-8

4 0
3 years ago
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