Answer:
The mass percentage of bromine in the original compound is 81,12%
Explanation:
<u>Step 1: Calculate moles AgBr</u>
moles AgBr = mass AgBr / molar mass AgBr
= 0.8878 g / 187.77 g/mol
= 0.00472812 moles AgBr
⇒
Since 1 mol AgBr contains 1 mol Br-
Then the amount of moles Br- in the original sample must also have been 0.00472812 moles
<u>Step 2:</u> Calculating mass Br-
mass Br- = molar mass Br x moles Br-
= 79.904 g/mol x 0.00472812 mol
= 0.377796 g Br-
⇒
There were 0.377796 g Br- in the original sample
<u>Step 3:</u> Calculating mass percentage Br-
⇒mass percentage = actual mass Br- / total mass x 100%
% mass Br = 0.377796 g / 0.4657 g x 100 %
= 81.12%
Answer: an addition reaction is one where two molecules react together to produce one. in this case a bromine molecule reacts with ethene and one of the bromine atoms is added to ethene
Explanation:
hope this helps :)
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