Answer:
Explanation:
a )
CH₃CH₂CH₂CH₂CH₂Br + KOH ⇒ CH₃CH₂CH₂CH₂CH₂OH
CH₃CH₂CH₂CH₂CH₂OH + acidic potassium dichromate ⇒ CH₃CH₂CH₂CH₂COOH
b )
CH₃CH₂CH₂CH₂CH₂Br + KCN ⇒ CH₃CH₂CH₂CH₂CH₂CN Hydrolysis ⇒ CH₃CH₂CH₂CH₂CH₂COOH .
c )
CH₃CH₂CH₂CH₂CH₂Br + KOH ⇒ CH₃CH₂CH₂CH₂CH₂OH
CH₃CH₂CH₂CH₂CH₂OH + acidic potassium dichromate ⇒ CH₃CH₂CH₂CH₂COOH + SOCl₂ ( thionyl chloride ) ⇒ CH₃CH₂CH₂CH₂COCl
d )
CH₃CH₂CH₂CH₂CH₂Br + KCN ⇒ CH₃CH₂CH₂CH₂CH₂CN Hydrolysis ⇒ CH₃CH₂CH₂CH₂CH₂COOH + PCC ( NH₃ ) ⇒ CH₃CH₂CH₂CH₂CH₂CONH₂
e )
CH₃CH₂CH₂CH₂CH₂Br + KCN ⇒ CH₃CH₂CH₂CH₂CH₂CN Hydrolysis ⇒ CH₃CH₂CH₂CH₂CH₂COOH + C₂H₅OH ( Ethyl alcohol + H⁺ )⇒
CH₃CH₂CH₂CH₂CH₂COOC₂H₅ ( ethyl hexanoate )
Answer:
Empirical formula of compound is C₄H₈O
Explanation:
Given data:
Mass of compound = 5.60 g
Mass of CO₂ = 13.7 g
Mass of H₂O = 5.60 g
Empirical formula of compound = ?
Solution:
Percentage of C:
13.7 g/5.60 g × 12/44× 100
2.45×0.273× 100 = 66.9%
Percentage of H:
5.60 g/ 5.60 g × 2.016/18 × 100
11.2%
Percentage of O:
(66.9% + 11.2%) - 100 = 21.9%
Grams atom of C , H, O
66.9/12 = 5.6
11.2 / 1.008 = 11.11
21.9 / 16 = 1.4
Atomic ratio:
C : H : O
5.6/1.4 : 11.11/1.4 : 1.4/1.4
4 : 8 : 1
Empirical formula:
C₄H₈O
The NH3, as it was the acid in the beginning and donated a proton to become more basic.
Answer:
1. c 2. b that's what i think i learned this in middle school i dont completely remember