the percent yield of the reaction is 100%.
The percent yield is calculated as the experimental yield divided by the theoretical yield x 100%:
% yield = actual yield / theoretical yield * 100%
% yield of a reaction in this case Rate
In this case, the molar mass of NaBr is 102.9 g / mol, as you know:
444 actual yield = 7.08 mol x 102.9 g / mol = 728.532 g
theoretical yield = 7.08 mol x 102.9 g / mol = 728.532 g
, Replaced by the definition of percent yield:
percent yield = 728.532 grams / 728.532 grams * 100%
percent yield = 100%
Finally, the percent yield of the reaction is 100%.
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FeBr3 is iron bromide. Also known as iron bromide. Iron bromide is an ionic compound in which iron is in a +3 oxidation state.
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Answer:
20 molecules of oxygen gas remains after the reaction.
Explanation:
Molecules of ethyne = 52
Molecules of oxygen gas = 150
According to reaction, 2 molecules of ethyne reacts with 5 molecules of oxygen gas.
Then 52 molecules of ethyne will react with:
of oxygen gas.
As we can see that we have 150 molecules of oxygen gas, but 52 molecules of ethyne will react with 130 molecules of oxygen gas. So, this means that ethyne is a limiting reagent and oxygen gas is an excessive reagent.
Remaining molecules of recessive reagent = 150 - 130 = 20
20 molecules of oxygen gas remains after the reaction.
The answer should be D, cells can only create an identical copy of the original cell.
(1) IS THE BALANCED EQUATION AND REACTANT-> MG AND N2 AND PRODUCT ->MGN2.
(2) IS NOT BALANCED AND REACTANT->CF4 AND BR AND PRODUCT IS CBR4 AND F2
