<u>Answer:</u> The partial pressure of carbon dioxide at equilibrium is 0.0056 atm
<u>Explanation:</u>
The given chemical equation follows:
<u>Initial:</u> 4.00
<u>At eqllm:</u> 4.00-2x x x
The expression of 
for above reaction follows:
The partial pressure of pure solids and liquids are taken as 1 in the equilibrium constant expression.
We are given:
Putting values in above expression, we get:
Neglecting the value of x = 718.28 because equilibrium pressure cannot be greater than initial pressure
Partial pressure of 
= 0.0056 atm
Hence, the partial pressure of carbon dioxide at equilibrium is 0.0056 atm
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Llowing certain substances to pass through it but not others, especially allowing the passage of a solvent but not of certain solutes.
Answer:
166 g/mol
Explanation:
Step 1: Write the neutralization reaction
H₂A + 2 NaOH ⇒ Na₂A + 2 H₂O
Step 2: Calculate the reacting moles of NaOH
48.3 mL of 0.0700 M NaOH react.
0.0483 L × 0.0700 mol/L = 3.38 × 10⁻³ mol
Step 3: Calculate the reacting moles of H₂A
The molar ratio of H₂A to NaOH is 1:2. The reacting moles of H₂A are 1/2 × 3.38 × 10⁻³ mol = 1.69 × 10⁻³ mol.
Step 4: Calculate the molar mass of H₂A
1.69 × 10⁻³ moles of H₂A have a mass of 0.281 g. The molar mass of H₂A is:
M = 0.281 g / 1.69 × 10⁻³ mol = 166 g/mol
