Answer:
A. The pressure will increase 4 times. P₂ = 4 P₁
B. The pressure will decrease to half its value. P₂ = 0.5 P₁
C. The pressure will decrease to half its value. P₂ = 0.5 P₁
Explanation:
Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.
<em>What would happen to the gas pressure inside the cylinder if you do the following?</em>
<em />
<em>Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.</em>
V₂ = 0.25 V₁. According to Boyle's law,
P₁ . V₁ = P₂ . V₂
P₁ . V₁ = P₂ . 0.25 V₁
P₁ = P₂ . 0.25
P₂ = 4 P₁
<em>Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.</em>
T₂ = 0.5 T₁. According to Gay-Lussac's law,

<em>Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.</em>
n₂ = 0.5 n₁.
P₁ in terms of the ideal gas equation is:

P₂ in terms of the ideal gas equation is:

Answer:
I Don't Know.
Explanation: I'm Sorry, But I'm Not Good At This Subject. Atleast I'm Being Honest.
Answer:
39.6 mL
Explanation:
Step 1: Write the balanced neutralization reaction
Ba(OH)₂(aq) + 2 CH₃COOH(aq) ⟶ Ba(CH₃COO)₂(aq) + 2 H₂O(l)
Step 2: Calculate the moles corresponding to 2.78 g of CH₃COOH
The molar mass of CH₃COOH is 60.05 g/mol.
2.78 g × 1 mol/60.05 g = 0.0463 mol
Step 3: Calculate the moles of Ba(OH)₂ needed to react with 0.0463 moles of CH₃COOH
The molar ratio of Ba(OH)₂ to CH₃COOH is 1:2. The moles of Ba(OH)₂ needed are 1/2 × 0.0463 mol = 0.0232 mol.
Step 4: Calculate the volume of 0.586 M solution that contains 0.0232 moles of Ba(OH)₂
0.0232 mol × 1 L/0.586 mol = 0.0396 L = 39.6 mL