the answer is fresh water
Answer:
CH₃CH₂CH₂COOH > CH₃CH₂COOH > ClCH₂CH₂COOH > ClCH₂COOH
Explanation:
Electron-withdrawing groups (EWGs) increase acidity by inductive removal of electrons from the carboxyl group.
Electron-donating groups (EDGs) decrease acidity by inductive donation of electrons to the carboxyl group.
- The closer the substituent is to the carboxyl group, the greater is its effect.
- The more substituents, the greater the effect.
- The effect tails off rapidly and is almost zero after about three C-C bonds.
CH₃CH₂-CH₂COOH — EDG — weakest — pKₐ = 4.82
CH₃-CH₂COOH — reference — pKₐ = 4.75
ClCH₂-CH₂COOH — EWG on β-carbon— stronger — pKₐ = 4.00
ClCH₂COOH — EWG on α-carbon — strongest — pKₐ = 2.87
the formula we is as follows:-
M1V1= M2V2
where
M1=1.2
V1=0.133l
V2=41l
M2=?
1.2 × 0.133 = 41 × M2
0.1596 = 41 × M2
M2 = 0.15960/41
M2 = 0.0038926829
Answer:
Option A:
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)
Explanation:
The half reactions given are:
Zn(s) → Zn^(2+)(aq) + 2e^(-)
Cu^(2+) (aq) + 2e^(-) → Cu(s)
From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).
While in the second half reaction, Cu^(2+) is reduced to Cu.
Thus, for the overall reaction, we will add both half reactions to get;
Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)
2e^(-) will cancel out to give us;
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)
