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Olin [163]
2 years ago
13

What happens when aluminum fills its valence shell?

Chemistry
2 answers:
maksim [4K]2 years ago
7 0
Aluminum is a metal and its oxidation state is +3. That means that it may lose three electrons and form cation Al +3. So the correct answer is option A. threee electrons are lost creating Al +3.

You can base your answer in the fact is in the group 13, that means that it has 3 electrons in its valence shell. Those are the three electrons that it can lose to form Al +3.

Also, you can search for the atomic number and do its electron configuration.

Z Al = 13 => 1s2 2s2 2p6 3s2 3p1. So, the electrons 3s2 3p1 are the three electrons that it can lose to form Al +3.
anzhelika [568]2 years ago
3 0

Three electrons are lost, creating AL+3! (So the correct answer is A.)

I can confirm this because I took the test! :D

Hope that helped!

Cheers, mate!

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Which equation represents a reduction half-reaction?
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Answer:

Fe3+ + 3e– --------> Fe

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Consider the redox reaction half equation;

Fe3+ + 3e– ------>Fe

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The brown gas NO2 and the colorless gas N2O4 exist in equilibrium,2NO2 N2O4.In an experiment, 0.625 mole of N2O4 was introduced
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Answer : The value of K'_c for the reaction is, 7.5

Explanation :

Initial concentration of N_2O_4 = \frac{Moles}{Volume}=\frac{0.625}{5}=0.125M

The balanced decomposition equilibrium reaction is,

                      N_2O_4(g)\rightleftharpoons 2NO_2(g)

Initial conc.     0.125          0

At eqm.          (0.125-x)       2x

The expression of equilibrium constant for the reaction will be:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

Now put all the values in this expression, we get :

K_c=\frac{(2x)^2}{(0.125-x)}

The concentration of N_2O4 at equilibrium = 0.0750 M

So, 0.125 - x = 0.0750

x = 0.05 M

Now put the value of 'x' in the above expression, we get:

K_c=\frac{(2\times 0.05)^2}{(0.0750)}

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Now we have to calculate the value of K_c for the given reaction.

2NO_2(g)\rightleftharpoons N_2O_4(g)   K'_c

As we know that,

K'_c=\frac{1}{K_c}

So,

K'_c=\frac{1}{0.133}

K'_c=7.5

Therefore, the value of K'_c for the reaction is, 7.5

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