Answer:
The answer to your question is: yield = 56.27%
Explanation:
Data
CH3CH2CH2CH2OH (l) → CH3 CH2CH2CH2Br
18.54 ml 1-butanol 15.65 g of 1-bromobutane
% yield = ?
density = 0.81 g/ml
MM = 74 g 1- butanol
MM = 137 g 1-bromobutane
Process
Calculate mass of 1- butanol
density = mass/volume
mass = density x volume
mass = 0.81 x 18.54
mass = 15.02 g of 1-butanol
Theoretical yield
74 g of 1- butanol ----------------- 137 g of 1-bromobutane
15.02 g of 1- butanol ------------- x
x = (15.02 x 137) / 74
x = 27.81 g of 1-bromobutane
% yield = experimental yield / theoretical yield x 100
% yield = 15.65 / 27.81 x 100
% yield = 56.28
Answer:
The answer to your question is: 30 g
Explanation:
Data
mass of water = 25 g
mass of salt = 5 g
Process
total mass = mass of water + mass of salt
= 25 + 5
= 30 g
62.23 = 1512.5001499999998 moles
Answer:
water supply
Explanation:
A drought is a natural thing where water is dried out
brainliest pls
(1) Ploar covalent. is the answer.
