<u>Answer:</u> The entropy change of the liquid water is 63.4 J/K
<u>Explanation:</u>
To calculate the entropy change for same phase at different temperature, we use the equation:

where,
= Entropy change
= molar heat capacity of liquid water = 75.38 J/mol.K
n = number of moles of liquid water = 3 moles
= final temperature = ![95^oC=[95+273]K=368K](https://tex.z-dn.net/?f=95%5EoC%3D%5B95%2B273%5DK%3D368K)
= initial temperature = ![5^oC=[5+273]K=278K](https://tex.z-dn.net/?f=5%5EoC%3D%5B5%2B273%5DK%3D278K)
Putting values in above equation, we get:

Hence, the entropy change of the liquid water is 63.4 J/K
HNO₃ + H₂S → S + NO + H₂<span>O
Assign Oxidation Number:
L.H.S R.H.S
N in HNO</span>₃ = +5 +2 = N in NO
S in H₂S = -2 0 = S in S
Write Half cell Reactions:
Reduction Reaction:
3e⁻ + HNO₃ → NO -------(1)
Oxidation Reaction:
H₂S → S + 2e⁻ -------(2)
Multiply eq. 1 with 2 and eq. 2 with 3 to balance electrons.
6e⁻ + 2 HNO₃ → 2 NO
3 H₂S → 3 S + 6e⁻
Cancel e⁻s,
______________________________
2 HNO₃ + 3 H₂S → 2 NO + 3 S + H₂O
Balance Oxygen Atoms by multiplying H₂O with 4, Hydrogen atoms will automatically get balance.
2 HNO₃ + 3 H₂S → 2 NO + 3 S + 4H₂O
Answer:
pH of buffer =4.75
Explanation:
The pH of buffer solution is calculated using Henderson Hassalbalch's equation:
![pH=pKa+log[\frac{[salt]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%5B%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
Given:
pKa = 3.75
concentration of acid = concentration of formic acid = 1 M
concentration of salt = concentration of sodium formate = 10 M
![pH=3.75+log[\frac{10}{1}]=3.75+1=4.75](https://tex.z-dn.net/?f=pH%3D3.75%2Blog%5B%5Cfrac%7B10%7D%7B1%7D%5D%3D3.75%2B1%3D4.75)
pH of buffer =4.75
Atoms of oxygen are electronegative and attract the shared electrons in their covalent bonds.