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meriva
2 years ago
11

A 3.50 amp power supply is used to deposit chromium from a solution of CrCl3. How long will it take to deposit 100.0 grams of ch

romium?
Chemistry
1 answer:
Norma-Jean [14]2 years ago
3 0

Explanation:

The given data is as follows.

   Current (I) = 3.50 amp,        Mass deposited = 100.0 g

  Molar mass of Cr = 52 g

It is known that 1 faraday of electricity will deposit 1 mole of chromium. As 1 faraday means 96500 C and 1 mole of Cr means 52 g.

Therefore, 100 g of Cr will be deposited by "z" grams of electricity.

                z \times 52 g = 96500 \times 100 g

                         z = \frac{96500 \times 100 g}{52 g}

                            = 185576.9 C

As we know that, Q = I × t

Hence, putting the given values into the above equation as follows.

                      Q = I × t

           185576.9 C = 3.50 amp \times t  

                      t = 53021.9 sec

Thus, we can conclude that 100 g of Cr will be deposited in 53021.9 sec.

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Answer :

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Li_2SO_4(aq)+Pb(CH_3COOH)_2(aq)\rightarrow 2LiCH_3COOH(aq)+PbSO_4(s)

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Molecular equation : It is defined as a balanced chemical equation where the ionic compounds are expressed in the form of molecules rather than component of ions.

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The insoluble salt that settle down in the solution is known an precipitate.

Part A  : potassium carbonate and lead(II) nitrate

The balanced molecular equation will be:

K_2CO_3(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbCO_3(s)

In this reaction, lead carbonate is an insoluble salt and potassium nitrate is a soluble solution.

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The balanced molecular equation will be:

Li_2SO_4(aq)+Pb(CH_3COOH)_2(aq)\rightarrow 2LiCH_3COOH(aq)+PbSO_4(s)

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3 years ago
How many kilograms of solvent would contain 0.43 mol of CaO in a 2.5 molality solution
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<h3>Answer:</h3>

0.024 kg CaO

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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  2. Parenthesis
  3. Exponents
  4. Multiplication
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<u>Step 1: Define</u>

0.41 mol CaO

2.5 M Solution

<u>Step 2: Identify Conversions</u>

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