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Colt1911 [192]
2 years ago
8

AICI, + NaNaCl + Al Did Al change oxidation number? yes No

Chemistry
1 answer:
mezya [45]2 years ago
3 0

<u>Answer:</u>

<em>Yes, Al change oxidation number</em>

<em></em>

<u>Explanation:</u>

AlCl_3  +Na >NaCl+Al

For a free element, oxidation number is 0.

Na and Al are free elements we find in the equation as they are not in combination with other elements.

For a compound, the sum of oxidation number of elements = 0.

We know oxidation number of chlorine is -1.

So let us find the oxidation number of Al in AlCl_3

Let x be the oxidation number of Al so,

AlCl_3=x+(3\times-1)=0\\\\x-3=0\\\\ x= +3

Oxidation number of Al in AlCl_3  is +3

Oxidation number of Al  changes from +3 to 0.

Decrease in oxidation number indicates AlCl_3 is getting reduced in the reaction.

Na oxidation number increases from 0 to +1 shows Na is getting oxidised in the reaction  

The change in oxidation number of the elements represents a redox reaction where both oxidation and reduction takes place simultaneously.

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2 years ago
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Answer:  " 13.5 g Al " ;

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<u>____________________________</u>

Explanation:

<u>____________________________</u>

<u>Note</u>: What is missing from the question is the "balanced chemical equation" for the "chemical reaction" that contains:

 The reactants:  "aluminum (Al) " ;  and "chlorine (Cl) " ;  and:

 The product:    "aluminum choloride (AlCl₃) " .

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The "balanced chemical equation" is:

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        2 Al   +   3 Cl₂   →   2 AlCl₃   ;

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<u>Note</u>: The molecular weight of "aluminum (Al)" is:   " 26.98 g /mol " .

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So:  We call solve using a technique known as:  "dimensional analysis" :

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  0.500 mol AlCl₃ * (\frac{2mol Al}{2mol AlCl_{3} }) * (\frac{26.98g Al}{1 mol Al}) = ?

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<u>Note</u>:  The units of "mol AlCl₃" cancel out to "1' ; and:

          The  units of "mol Al" cancel out to "1" ; and we are left with:

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 " \frac{(0.500 * 2 * 26.98)}{2}   g Al ["grams of aluminum"] ;

____________________________

<u>Note</u>: We can "cancel out the "2's" ; since "2/2 = 1 " ; and we have:

 →  (0.500 * 26.98) g Al ;

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         →  Round to 3 (Three) significant figures;

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   =  13.5 g Al ; that is:  "13.5 grams of aluminum."

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 Hope this is helpful!  

      Best wishes to you in your academic pursuits—and within the "Brainly" community!

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3 years ago
The half-life of cesium-137 is 30 years. suppose we have a 200-mg sample.
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