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%
Answer:
The entropy decreases.
Explanation:
The change in the standard entropy of a reaction (ΔS°rxn) is related to the change in the number of gaseous moles (Δngas), where
Δngas = n(gaseous products) - n(gaseous reactants)
- If Δngas > 0, the entropy increases
- If Δngas < 0, the entropy decreases.
- If Δngas = 0, there is little or no change in the entropy.
Let's consider the following reaction.
2 H₂(g) + O₂(g) ⟶ 2 H₂O(l)
Δngas = 0 - 3 = -3, so the entropy decreases.
The reaction that takes place in a nuclear fission reactor is as follows: 235/92 U + 1/0n 94/36Kr + 139/56 Ba + 3/0n.
<h3>What is a nuclear fission reactor?</h3>
A nuclear fission reactor is the place where nuclear chain reactions occur that produce energy by fission.
Nuclear fission is the nuclear reaction in which a large nucleus splits into smaller ones with the simultaneous release of energy.
Therefore, the option that involves the splitting of atoms into smaller ones is as follows: 235/92 U + 1/0n 94/36Kr + 139/56 Ba + 3/0n.
Learn more about nuclear fission reactor at: brainly.com/question/10203508
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Answer:
A) Ionic bond
___transfer of electrons
B) Covalent bond
___sharing of electrons
C) Metallic bond
___freely moving electrons
Answer: 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether
Explanation:
First we have to calculate the moles of diethyl ether
As, 1 mole of diethyl ether require heat = 26.5 kJ
So, 1.01 moles of diethyl ether require heat = 
Thus 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether
