Answer:
The answer is "Option D".
Explanation:
The behavior of 0.1M NaCl also isn't substantially larger objectively than those of 0.05M NaCl because a p-value above 0.05 (p>0.05) indicates no ability to tell differential and is a strong proof in favor of a null hypothesis.
The other wrong choices can be defined as follows:
- Option A as it's just the reverse of the correct answer to the null.
- Options B and C because p worth tests to support nor oppose the null hypothesis.
Answer:
Faraday's constant will be smaller than it is supposed to be.
Explanation:
If the copper anode was not completely dry when its mass was measured, mass of the copper must be heavier than it should have been. Hence, the calculated Faraday’s constant would be smaller than it is supposed to be since when calculating Faraday’s Constant, the charge transferred is divided by the moles of electrons.
Answer:
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Answer:
O₂; KCl; 33.3
Explanation:
We are given the moles of two reactants, so this is a limiting reactant problem.
We know that we will need moles, so, lets assemble all the data in one place.
2KCl + 3O₂ ⟶ 2KClO₃
n/mol: 100.0 100.0
1. Identify the limiting reactant
(a) Calculate the moles of KClO₃ that can be formed from each reactant
(i)From KCl
(ii) From O₂
O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.
KClO₃ is the excess reactant.
2. Moles of KCl left over
(a) Moles of KCl used
(b) Moles of KCl left over
n = 100.0 mol - 66.67 mol = 33.3 mol
