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 statement is true.....
Answer:
Approx. 20 moles of iron.
Explanation:
There are approx. 6.022 x 10.23 iron atoms per mole of iron. In this quantity, The number of iron atoms has a mass of
55.85 ⋅ g.
Answer:
residue is whatever remains after something else has been removed while filtrate is the liquid or solution that has passed through a filter, and which has been separated from the filtride.
filtration id done by placing a filter paper on the beaker or container then pour the filtride then let it settle and it will pass through and you will have the fitrate and residue
Explanation:
Answer:
five electrons
Bromine (Z=35), which has 35 electrons, can be found in Period 4, Group VII of the periodic table. Since bromine has 7 valence electrons, the 4s orbital will be completely filled with 2 electrons, and the remaining five electrons will occupy the 4p orbital
