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1 year ago

6.

Chemistry

1 answer:

RUDIKE [14]1 year ago

3 0

Answer:

THE CURRENT REQUIRED TO PRODUCE 193000 C OF ELECTRICITY IS 35.74 A.

Explanation:

Equation:

Al3+ + 3e- -------> Al

3 F of electricity is required to produce 1 mole of Al

3 F of electricity = 27 g of Al

If 18 g of aluminium was used, the quantity of electricity to be used up will be:

27 g of AL = 3 * 96500 C

18 G of Al = x C

x C = ( 3 * 96500 * 18 / 27)

x C = 193 000 C

For 18 g of Al to be produced, 193000 C of electricity is required.

To calculate the current required to produce 193 000 C quantity of electricity, we use:

Q = I t

Quantity of electricity = Current * time

193 00 = I * 1.50 * 60 * 60 seconds

I = 193 000 / 1.50 * 60 *60

I = 193 000 / 5400

I = 35.74 A

The cuurent required to produce 193,000 C of electricity by 18 g of aluminium is 35.74 A

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Answer:

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Explanation:

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Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:

CuF₂ ⇄ Cu²⁺ + 2F⁻

At t=0 x - 0.20

At t =equilibrium (x-s') s' 0.20+2s'

The expression for Solubility product for CuF₂ is:

$K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2$

$1.6209\times 10^{-6}={s}'\times ({0.20+2{s}'})^2$

Solving for s', we get

s' = 4.0×10⁻⁵ M

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.

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