Answer:
The question is incomplete as some details are missing. Here is the complete question ; A chemist adds 45.0mL of a 0.434M copper(II) sulfate CuSO4 solution to a reaction flask. Calculate the mass in grams of copper(II) sulfate the chemist has added to the flask. Round your answer to 2 significant digits
Explanation:
The step by step explanation is as shown in the attachment
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:
Yes
Explanation:
Because it was false because it has been the substance abuse permalink
I think the answer is:
B. Chemical Change.
Answer:
atomic mass of X is 48.0 amu
Explanation:
Let y be the atomic mass of X
Molar mass of O_2 is = 2×16 = 32 g / mol
X + O2 -----> XO_2
According to the equation ,
y g of X reacts with 32 g of O_2
24 g of X reacts with Z g of O_2
Z = ( 32×24) / y
But given that 24.0 g of X exactly reacts with 16.0 g of O_2
So Z = 16.0
⇒ (32×24) / y = 16.0
⇒ y = (32×24) / 16
y= 48.0
So atomic mass of X is 48.0 amu