Answer:
12.09 L
Explanation:
Step 1: Convert 826.1 mmHg to atm
We will use the conversion factor 760 mmHg = 1 atm.
826.1 mmHg × 1 atm/760 mmHg = 1.087 atm
Step 2: Convert 427.8 J to L.atm
We will use the conversion factor 101.3 J = 1 L.atm.
427.8 J × 1 L.atm/101.3 J = 4.223 L.atm
Step 3: Calculate the change in the volume
Assuming the work done (w) is 4.223 L.atm against a pressure (P) of 1.087 atm, the change in the volume is:
w = P × ΔV
ΔV = w/P
ΔV = 4.223 L.atm/1.087 atm = 3.885 L
Step 4: Calculate the final volume
V₂ = V₁ + ΔV
V₂ = 8.20 L + 3.885 L = 12.09 L
Answer : The correct option is, (b) +115 J/mol.K
Explanation :
Formula used :

where,
= change in entropy
= change in enthalpy of vaporization = 40.5 kJ/mol
= boiling point temperature = 352 K
Now put all the given values in the above formula, we get:



Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K
Just search it up on the internet:)
Answer:
0.185M sulfuric acid
Explanation:
Based on the reaction:
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
<em>1 mole of sulfuric acid reacts with 2 moles of KOH</em>
Initial moles of H₂SO₄ and KOH are:
H₂SO₄: 0.750L ₓ (0.470mol / L) = <em>0.3525 moles of H₂SO₄</em>
KOH: 0.700L ₓ (0.240mol / L) = <em>0.168 moles of KOH</em>
The moles of sulfuric acis that react with KOH are:
0.168mol KOH ₓ (1 mole H₂SO₄ / 2 moles KOH) = 0.0840 moles of sulfuric acid.
Thus, moles that remain are:
0.3525moles - 0.0840 moles = <em>0.2685 moles of sulfuric acid remains</em>
As total volume is 0.700L + 0.750L = 1.450L, concentration is:
0.2685mol / 1.450L = <em>0.185M sulfuric acid</em>
Answer:
156.4g K
Explanation:
I'm not sure if it is correct but I think it should be this
What do we know so far?: 2K + 1Cl2 -> 2KCl, 2 mol of Cl2
What are we looking for?: #g of K
What is the ratio of K to Cl2?: 2:1
Set up equation: 2molCl2 x 
Cancel unwanted units: 2 x 
Answer we got: 2 x 2mol K = 4mol K
Converting moles to grams: 4 x 39.1 (molar mass of K) = 156.4g K