The answer is B plastic bags made from petroleum.
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
2..............................................
Answer:
balanced in ACID not BASE
Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O
Answer
Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O
Explanation:
Cr2O7^2-(aq) + Hg(l) ----> Hg^2+(aqH) + Cr^3+(aq)
add H^1+ (acid) to capture the O and make 7 water molecules
Cr2O7^2-(aq) + Hg(l) + H^1+ ----> Hg^2+(aqH) + Cr^3+(aq) + 7H2O
Cr goes from +6 to +3 by gaining 3 e
Hg goes from 0 to +2 by losing 2 e
we need 3 Hg for every 2 Cr
so
Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O
2 Cr on the right and left
Net 12 positive charges on the right and the left
3 Hg on the right and left
14 H on the right and left
the equation is balanced
we cannot balance the equation in a basic solution with OH^1-
we have plenty of O in the dichromate ion. we need to convert it to water which take free H^1+ from the acid
No chemical because, its changeing the stuff
