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Colt1911 [192]
3 years ago
8

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

Chemistry
1 answer:
mezya [45]3 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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A gas occupies a volume of 0.7 L at 10.1 kPa. What volume will the gas occupy at 101 kPa?
Nataly [62]

Answer:

7L

Explanation:

Divide 101 by 10.1(kPa) and you get 10.

10 x .7 = 7

4 0
2 years ago
How many grams of 2.5% "h" cream should be mixed with360 grams of 0.25% "h" cream to make a 1% "h" cream?
Leona [35]
Answer is: 180 <span>grams of 2.5% "h" cream should be mixed.
</span>ω₁ = 2.5% ÷ 100% = 0.025.
ω₂ = 0.25% ÷ 100% = 0.0025.
ω₃ = 1% ÷ 100% = 0.01.
m₂ = 360 g.
m₃ = m₁ + 360 g.
m₁ = ?.
ω₁ · m₁ + ω₂ · m₂ = ω₃ · m₃.
0.025 · m₁ + 0.0025 · 360 g = 0.01 · (m₁ + 360 g).
0.025 · m₁ + 0.9 g = 0.01 · m₁ + 3.6 g.
0.015 · m₁ = 2.7 g.
m₁ = 180 g.
4 0
3 years ago
(a) Consider the reaction of hydrogen sulfide with methane, given below:
Burka [1]

Answer:

a. H2S(g)/t = 1.48 mol/s

CS2(g)/t = 0.740mol/s

H2(g)/t = 2.96mol/s

b. Ptot /t = 981torr/min

Explanation:

a. Based on the reaction:

CH4(g) + 2 H2S(g) → CS2(g) + 4 H2(g)

<em>1 mole of CH4 reacts with 2 moles of H2S producing 1 mole of CS2 and 4 moles of 4H2</em>

<em />

If CH4 decreases at the rate of 0.740mol/s, H2S decreases twice faster, that is 0.740mol/s = 1.48 mol/s

CS2 is produced with the same rate of CH4 because 1 mole of CH4 produce 1 mole of CS2 = 0.740mol/s

The H2 is produced four times faster than CH4 is decreased, that is:

0.740mol/s * 4 = 2.96mol/s

b. With the reaction:

2 NH3(g) → N2(g) + 3 H2(g)

2 moles of ammonia are consumed whereas 1 mole of N2 and 3 moles of H2 are produced.

That means 2 moles of gas are consumed and 4 moles of gas are produced.

If the NH3 decreases at a rate of 327torr/min, the gases are produced in a rate twice faster. That is 327torr/min*2 =

654torr/min

The rate of change of the total pressure is rate of reactants + rate of products:

654torr/min + 327torr/min =

981torr/min

6 0
4 years ago
List the procedural steps, from start to finish, that are required to convert 2‑naphthol and allyl bromide into allyl 2‑naphthyl
elena-14-01-66 [18.8K]
<h3>Procedural steps are: - </h3>
  1. On a small scale, the reaction is carried out by combining the alcohol, the haloalkane, and the phase transfer catalyst in a conical vial.
  2. To start the reaction, sodium hydoxide (base) is added.
  3. To prevent solvent evaporation, the reaction flask is covered and stirred during the reaction.
  4. TLC monitors the reaction's progress to ensure that no time is wasted.
  5. To remove any remaining water, the reaction solution is dried over calcium chloride.
  6. Column chromatography is used to purify the product, and evaporation is used to collect it.
<h3>What is Catalysis?</h3>

Catalysis is the process of boosting the pace of a chemical reaction by using a catalyst. Catalysts are not consumed in the reaction and so survive it.

To learn more about catalysis from the given link

brainly.com/question/1372992

#SPJ4

8 0
2 years ago
25.00 mL of a H2SO4 solution with an unknown concentration was titrated to a phenolphthalein endpoint with 28.11 mL of a 0.1311
LuckyWell [14K]

Answer:

Concentration of the H₂SO₄ solution is 0.0737 M

Explanation:

Equation of the neutralization reaction between the acid, H₂SO₄, and the base, NaOH, is given below:

H₂SO₄ + 2NaOH -----> Na₂SO₄ + 2H₂O

From the above equation, one mole of acid requires 2 moles of base for complete neutralization which occurs at phenolphthalein endpoint.

mole ratio of acid to base, nA/nB = 1:2

Concentration of the base, Cb = 0.1311 M

Volume of base, Vb, = 28.11 mL

Concentration of acid, Ca = ?

Volume of acid, Va + 25.0 mL

Using the formula, CaVa/CbVb = nA/nB

making Ca subject of the formula, Ca = Cb*Vb*nA/Va*nB

substituting the values into the equation

Ca = (0.1311 * 28.11 * 1) / 25.0 * 2 = 0.0737 M

Therefore, concentration of the H₂SO₄ solution is 0.0737 M

5 0
4 years ago
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