True.......................................
Answer:
The percent by mass of copper in the mixture was 32%
Explanation:
The ammount of HNO₃ used is:
mol HNO₃ = volume * concentration
mol HNO₃ = 0.015 l * 15.8 mol/l
mol HNO₃ = 0.237 mol
According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.
Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.
1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:
0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.
Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:
100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g
Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%
The value of the given expression up to correct significant figures is
.
- The rule that applies for the multiplication and division is :
The least number of significant figures in any number of the problem determines the number of significant figures in the answer.
- The rule that applies for the addition and subtraction is :
The least precise number present after the decimal point determines the number of significant figures in the answer.
Given:
The expression: (23.4m)(14m)
To find:
The answer up to correct a number of significant figures.
Solution
Number of significant figures in '23.4' = 3
Number of significant figures in '14' = 22

The number with the least significant figures is '14' is 2, so the answer will be reported up to 2 significant figures:

Number of significant figures in '330' = 3
The value of the given expression up to correct significant figures is
.
Learn more about significant figures here:
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It’s 1 hope this helps :)
Sorry this is late but it would be the air mass
Hope this halpes :))