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2 years ago

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

Chemistry

1 answer:

mezya [45]2 years ago

3 0

Answer:

Yes, Al change oxidation number

Explanation:

$AlCl_3 +Na >NaCl+Al$

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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HELP!!

irakobra [83]

Answer: In physics, potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors.

Explanation:

8 0

2 years ago

What mass of aluminum is needed to produce 0.500 mole of aluminum chloride?

3241004551 [841]

Answer: " 13.5 g Al " ;

→ that is: "13.5 grams of aluminum."

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Explanation:

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Note: 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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Note: 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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Note: 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"] ;

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

→ (0.500 * 26.98) g Al ;

= 13.49 g Al ;

→ Round to 3 (Three) significant figures;

→ Since: "0.500" has 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 0

3 years ago

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

Assoli18 [71]

a) To find the mass after t years:

we will use this formula:

A = Ao / 2^n

when A =the amount remaining

and Ao = the initial amount

and n = t / t(1/2)

by substitution:

∴ A = 200 mg/ 2^(t/30y)

b) Mass after 90 y :

by using the previous formula and substitute t by 90 y

A = 200mg/ 2^(90y/30y)

∴ A = 25 mg

C) Time for 1 mg remaining:

when A= Ao/ 2^(t/t(1/2)

so, by substitution:

1 mg = 200 mg / 2^(t/30y)

∴2^(t/30y) = 200 mg by solving for t

∴ t = 229 y