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iogann1982 [59]
1 year ago
6

A current of 5. 68 a is passed through a Fe(NO3)2 solution. How long, in hours, would this current have to be applied to plate o

ut 7. 20 g of iron?
Chemistry
1 answer:
daser333 [38]1 year ago
8 0

There are  1.2 hr would this current have to be applied to plate out 7. 20 g of iron .

Calculation ,

Given ; Current ( I ) = 5. 68 A

In Fe(NO_{3} )_{2} , the valancy of Fe is +2 .

2 moles of e^{-} are required for the decomposition of 1 mole of Fe .

7. 20 g  of Fe in moles  = 7. 20 g /55.845 g/mol =0.12 mole

x moles of  e^{-} are required for the decomposition of 0.128 mole of Fe .

moles of   e^{-} are required = 0.256 moles

Charge on 1 mole  of   e^{-} = 96500 C

Charge on 0.256  mole  of   e^{-} = 24704 C

Current ( I )= Q/t  

t =Q / I = 24704 C/5. 68 A = 4349 sec = 1.2 hr

Therefore , there are  1.2 hr would this current have to be applied to plate out 7. 20 g of iron .

To learn more about iron

brainly.com/question/18500540

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The question is incomplete, the complete question is:

A 7.00 L tank at 21.4^oC is filled with 5.43 g of sulfur hexafluoride gas and 14.2 g of sulfur tetrafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas. Round each of your answers to significant digits.

<u>Answer:</u> The mole fraction of sulfur hexafluoride is 0.221 and that of sulfur tetrafluoride is 0.779

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

  • <u>For sulfur hexafluoride:</u>

Given mass of sulfur hexafluoride = 5.43 g

Molar mass of sulfur hexafluoride = 146.06 g/mol

Putting values in equation 1, we get:

\text{Moles of sulfur hexafluoride}=\frac{5.43g}{146.06g/mol}=0.0372mol

  • <u>For sulfur tetrafluoride:</u>

Given mass of sulfur tetrafluoride = 14.2 g

Molar mass of sulfur tetrafluoride = 108.07 g/mol

Putting values in equation 1, we get:

\text{Moles of sulfur tetrafluoride }=\frac{14.2g}{108.07g/mol}=0.1314mol

Total moles of gas in the tank = [0.0372+ 0.1314] mol = 0.1686 mol

Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:

\chi_A=\frac{n_A}{n_A+n_B} .....(2)

where n is the number of moles

Putting values in equation 2, we get:

\chi_{SF_6}=\frac{0.0372}{0.1686}=0.221

\chi_{SF_4}=\frac{0.1314}{0.1686}=0.779

Hence, the mole fraction of sulfur hexafluoride is 0.221 and that of sulfur tetrafluoride is 0.779

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