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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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charle [14.2K]

Answer:

pH  = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH-   Ksp = 3X10^-15

Zn+2 + 4OH-   --> Zn(OH)4-2   Kf = 2X10^15

K = Ksp X Kf

  = 3*2*10^-15 * 10^15

  = 6

Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M

                Zn(OH)₂ + 2OH⁻(aq)  --> Zn(OH)₄²⁻(aq)

Initial:           0             0.3                      0

Change:                      -2x                     +x

Equilibrium:               0.3 - 2x                 x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²  

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

                                    = 0.3 -(2*0.089)

                                  = 0.122

pOH = -log[OH⁻]

         = -log 0.122

          = 0.91

pH = 14-0.91

     = 13.09

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The allotropic form of a non – metal is a good conductor of electricity. Name the nonmetal.
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A 110 g copper bowl contains 240 g of water, both at 21.0°C. A very hot 410 g copper cylinder is dropped into the water, causing
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Answer:

There is 98.76 kJ energy transfered to the water as heat.

Explanation:

<u>Step 1:</u> Data given

Mass of copper bowl = 110 grams

Mass of water = 240 grams

Temperature of water and copper = 21.0 °C

Mass of the hot copper cylinder = 410 grams

8.6 grams being converted to steam

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⇒ with ΔT = the change in temperature = T2 - T1 = 100 °C - 21.0 = 79°C

⇒ with mass of vapor = 8.60 grams = 0.0086 kg

⇒ with Lw = The latent heat of vaporization (water to steam) = 22.6 *10^5 J/kg

Q = 0.24kg * 4184 J/kg°C *79°C + 0.0086 kg*22.6*10^5 J/kg

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