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3 years ago

2AlCl3 --> 2Al + 3Al2

Chemistry

1 answer:

Eva8 [605]3 years ago

7 0

Answer:

1597,5 gm AlCl3 decomposed

Explanation:

2AlCl3 --> 2Al + 3Al2 If you produce 15 moles of Al, how many grams of AlCl3 decomposed?

FIRST, you need to write down the the correct reaction. Your equation makes no sense

here is the correct question

2AlCl3 --> 2Al + 3Cl2 If you produce 15 moles of Al, how many grams of AlCl3 decomposed?

for every mole of Al produced, we decomposed a mole of AlCl3

so we produced 15 moles of Al, so we decomposed 15 moles of AlCl3

15 moles are 15 X 106.5 = 1597,5 gm AlCl3 decomposed

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3 years ago

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Based on the following chemical equation how many hydrogen atoms are present in the products side?

e-lub [12.9K]

Answer:

the answer is 6

Explanation:

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3 years ago

A solution prepared by mixing 10 ml of 1 m hcl and 10 ml of 1.2 m naoh has a ph of

blagie [28]

Answer: pH of resulting solution will be 13

Explanation:

pH is the measure of acidity or alkalinity of a solution.

Moles of $H^+$ ion = $Molarity\times {\text {Volume in L}}=1M\times 0.01L=0.01mol$

Moles of $OH^-$ ion = $Molarity\times {\text {Volume in L}}=1.2M\times 0.01L=0.012mol$

$HCl+NaOH\rightarrow NaCl+H_2O$

For neutralization:

1 mole of $H^+$ ion will react with 1 mole of $OH^-$ ion

0.01 mol of $H^+$ ion will react with = $\frac{1}{1}\times 0.01mole$ of $OH^-$ ion

Thus (0.012-0.01)= 0.002 moles of $OH^-$ are left in 20 ml or 0.02 L of solution.

$[OH^-]=\frac{0.002}{0.02L}=0.1M$

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7 0

3 years ago

I am in desperate need of help will give 40 points to whoever helps me!!!!! I have a quiz tomorrow but I have no idea how to do

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All of the questions here are pertaining to the colligative properties of a solution and the preparation of solutions. Maybe, it would be best if you understand the equations to be used in order to answer these questions.
Freezing point depression or Boiling point elevation:

ΔT = -K (m) (i)

ΔT is the change in the freezing point or the boiling point not the freezing point/boiling point. Therefore, it should be added to the original value of the property of the solvent.

K is a constant called the molal freezing point depression constant and for the boiling point is the boiling point elevation constant. It is a property of the solvent.

m is the concentration of the solute in the solvent in terms of molality or kg solute/kg solvent.

i is the vant hoff factor which will represent the number of ions which the solute dissociates when in solution.

5 0

3 years ago

Enter your answer in the provided box.

Lina20 [59]

Answer:

5.06atm

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (Litres)

V2 = final volume (Litres)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

P1 = 1.34 atm

P2 = ?

V1 = 5.48 L

V2 = 1.32 L

T1 = 61 °C = 61 + 273 = 334K

T2 = 31 °C = 31 + 273 = 304K

Using P1V1/T1 = P2V2/T2

1.34 × 5.48/334 = P2 × 1.32/304

7.34/334 = 1.32P2/304

Cross multiply

334 × 1.32P2 = 304 × 7.34

440.88P2 = 2231.36

P2 = 2231.36/440.88

P2 = 5.06

The final pressure is 5.06atm

6 0

3 years ago