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o-na [289]
3 years ago
9

How many molecules of HCI would react with 2 moles of AI?

Chemistry
2 answers:
laila [671]3 years ago
7 0

Answer:

A:6.02+3H2

Explanation:

trasher [3.6K]3 years ago
5 0

Answer:

It is D. 1.806 x 1024    

Have a nice day <3

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A study is conducted to examine the effect of NaCl concentration on enzyme activity with the following results. Activity was tes
Marina86 [1]

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.
3 0
3 years ago
the copper anode is weighed before and after the electrolysis reation. if the copper anode is not completely dry when it is weig
Dmitrij [34]

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.

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3 years ago
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Answer:

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4 0
2 years ago
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Un catión trivalente contiene 75 nucleones y 45 neutrones, Señale el número de electrones que posee dicho catión.
FromTheMoon [43]

Answer:

10 electrons

Explanation:

5 0
3 years ago
Consider the following chemical reaction: 2KCl + 3O2 --&gt; 2KClO3. If you are given 100.0 moles of KCl and 100.0 moles of O2...
g100num [7]

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

\text{Moles of KClO}_{3} = \text{100.0 mol KCl} \times \dfrac{\text{2 mol KClO}_{3}}{\text{2 mol KCl}} = \text{100.0 mol KClO}_{3}

(ii) From O₂

\text{Moles of KClO}_{3} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KClO}_{3}}{\text{3 mol O}_{2}} = \text{66.67 mol KClO}_{3}

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

\text{Moles used} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KCl}}{\text{3 mol O}_{2}} = \text{66.67 mol KCl}

(b) Moles of KCl left over

n = 100.0 mol - 66.67 mol = 33.3 mol

3 0
3 years ago
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