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
Looks like it would be A................
The correct answer is: [C]: " mg " {"milligrams"} .
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Answer:
The first row fits in period 2
The second row fits in period 3
Explanation:
Elements arranged vertically in the periodic table in groups that share similar chemical properties.
Elements are also organized horizontally in rows or periods
The length of each period is determined by the number of electrons that can occupy the sublevels being filled in that period.
