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kow [346]
4 years ago
8

How many grams of lithium chloride can be produced from 14.3 g of lithium chlorate when it decomposes into lithium chloride and

oxygen gas?
Chemistry
1 answer:
vampirchik [111]4 years ago
3 0

6.7 grams of lithium chloride will be produced.

<h3><u>Explanation:</u></h3>

Lithium chlorate is LiClO₃ and lithium chloride is LiCl. The reaction is,

2LiClO₃ = 2LiCl +3O₂.

So here, 2 moles of lithium chlorate produces 2 moles of lithium chloride.

Or, one molecule of lithium chlorate will produce one mole of lithium chloride.

Molecular weight of lithium chlorate =6.9 + 35.5 + 16\times3 = 90.4.

So, 14.3 grams of lithium chlorate has 0.16 moles of lithium chlorate.

Thereby, moles of lithium chloride produced is 0.16 moles.

Molecular weight of lithium chloride = 6.9+35.5 = 42.4 grams.

So weight of lithium chloride produced = 42.4 \times 0.16 = 6.7 grams.

Thus, weight of lithium chloride produced will be 6.7 grams.

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Gwar [14]

Answer:

Explanation:

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6 0
3 years ago
If you want to use a serial dilution to make a 1/50 dilution. The first dilution you make is a 1/5 dilution with a total volume
uranmaximum [27]

Answer:

0.2 mL stock solution, 0.8 solvent, 0.1 mL first solution and 0.9 solvent

Explanation:

The final volume for fist solution is 1 mL and concentration must will be 1/5, then 1 mL/5=0.2 mL. For complete the 1 mL add the missing solvent volume 1 mL-0.2 mL=0.8 mL. For second solution, assuming final volume is 1 mL, and concentration 1/10, then we have 1 mL /10=0.1 mL solution 1/5. Completing volume, 1 mL-0.1 mL= 0.9 mL solvent.

7 0
3 years ago
658 mL of 0.250 M HCl solution is mixed with 325 mL of 0.600 M HCl solution. What is the molarity of the resulting solution
Tju [1.3M]

The molarity of the resulting solution obtained by mixing 658 mL of 0.250 M HCl solution with 325 mL of 0.600 M HCl solution is 0.366 M

We'll begin by calculating the number of mole of HCl in each solution. This can be obtained as follow:

<h3>For solution 1:</h3>

Volume = 658 mL = 658 / 1000 = 0.658 L

Molarity = 0.250 M

<h3>Mole of HCl =?</h3>

Mole = Molarity x Volume

Mole of HCl = 0.250 × 0.658

<h3>Mole of HCl = 0.1645 mole</h3>

<h3>For solution 2:</h3>

Volume = 325 mL = 325 / 1000 = 0.325 L

Molarity = 0.6 M

<h3>Mole of HCl =?</h3>

Mole = Molarity x Volume

Mole of HCl = 0.6 × 0.325

<h3>Mole of HCl = 0.195 mole</h3>

  • Next, we shall determine the total mole of HCl in the final solution. This can be obtained as follow:

Mole of HCl in solution 1 = 0.1645 mole

Mole of HCl in solution 2 = 0.195 mole

Total mole = 0.1645 + 0.195

<h3>Total mole = 0.3595 mole</h3>

  • Next, we shall determine the total volume of the final solution.

Volume of solution 1 = 0.658 L

Volume of solution 2 = 0.325 L

Total Volume = 0.658 + 0.325

<h3>Total Volume = 0.983 L</h3>

  • Finally, we shall determine the molarity of the resulting solution.

Total mole = 0.3595 mole

Total Volume = 0.983 L

<h3>Molarity =?</h3>

Molarity = mole / Volume

Molarity = 0.3595 / 0.983

<h3>Molarity = 0.366 M</h3>

Therefore, the molarity of the resulting solution is 0.366 M

Learn more: brainly.com/question/25342554

3 0
3 years ago
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Furkat [3]
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6 0
4 years ago
Calculate the time needed for a constant current of 0.961 a to deposit 0.500 g of co(ii) as
Vera_Pavlovna [14]

1.70 × 10³ seconds

<h3>Explanation </h3>

\text{Co}^{2+} + 2 e⁻ → \text{Co}

It takes two moles of electrons to reduce one mole of cobalt (II) ions and deposit one mole of cobalt.

Cobalt has an atomic mass of 58.933 g/mol. 0.500 grams of Co contains 0.500 / 58.933 = 8.484\times 10^{-3} \; \text{mol} of Co atoms. It would take 2 \times 8.484 \times 10^{-3} = 0.01697 \; \text{mol} of electrons to reduce cobalt (II) ions and produce the 8.484\times 10^{-3} \; \text{mol} of cobalt atoms.

Refer to the Faraday's constant, each mole of electrons has a charge of around 96 485 columbs. The 0.01697 mol of electrons will have a charge of 1.637 \times 10^{3} \; \text{C}. A current of 0.961 A delivers 0.961 C of charge in one single second. It will take 1.637 \times 10^{3} / 0.961 = 1.70 \times 10^{3} \; \text{s} to transfer all these charge and deposit 0.500 g of Co.

4 0
3 years ago
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