Answer:
b. 1.5 atm.
Explanation:
Hello!
In this case, since the undergoing chemical reaction suggests that two moles of A react with one moles of B to produce two moles of C, for the final pressure we can write:
Now, if we introduce the stoichiometry, and the change in the pressure 
we can write:
Nevertheless, since the reaction goes to completion, all A is consumed and there is a leftover of B, and that consumed A is:
Thus, the final pressure is:
Therefore the answer is b. 1.5 atm.
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Assuming that you’re looking for the concentration of water in the solution, then it would be 0.028 M.
You would have to use the formula:
c1v1 = c2v2, where c =concentration and
v = volume
C1 = ?
V1 = 250 mL
C2 = 0.2 M
V2 = 35 mL
C1 x 250 mL = 0.2 M x 35 mL
C1 = (0.2 M x 35 mL) / 250 mL
C1 = 0.028 M of water added to 35mL of 0.2M HCl
Therefore, there is 0.028 M of water added to 35mL of 0.2M HCl
Answer:
16,,24Mg 17,,a24.1 18a mass number of the most abundant isotope
Explanation:
atomic number of Mg is 12 ,therefore its mass number should be the value that is very close to 24.
24.1 is the value of thee most abundant isotope.
Answer: from the hotter surface to the colder one
Explanation:
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%
