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ioda
3 years ago
7

How many kilowatt-hours of electricity are used to produce 4.50 kg of magnesium in the electrolysis of molten mgcl2 with an appl

ied emf of 5.00 v?
Chemistry
1 answer:
ratelena [41]3 years ago
3 0
First, we need to determine the half reaction of magnesium. It would be expressed as:

Mg2+ + 2e- = Mg

Given the mass of magnesium metal that is produced, we calculate the total charge of the electrolysis by the relations we can get from the half reaction. We do as follows:

4.50 kg Mg ( 1000 g / 1 kg ) ( 1 mol / 24.305 g ) ( 2 mol e- / 1 mol Mg ) ( 96500 C / 1 mol e- ) = 35733388.2 C

We are given the applied EMF in units of V. This value is equal to J/C. So, 5 V is equal to 5 J/C.

35733388.2 C (5 J/C) = 178666941 J
178666941 J ( 1 kW-h / 3.6x10^6 J ) = 49.63 kW-h
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The possible valances can be determined by electron configuration and electron negativity

Good Luck even though this was asked 2 weeks ago

Explanation:

All atoms strive for stability. The optima electron configuration is the electron configuration of the VIII A family or inert gases.

Look at the electron configuration of the nonmetal and how many more electrons the nonmetal needs to achieve the stable electron configuration of the inert gases. Non metals tend to be negative in nature and gain electrons. ( They are oxidizing agents)

For example Florine atomic number 9 needs one more electron to reach a valance number of 8 electrons to equal Neon atomic number 10. Hence Flowrine has a valance of -1

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Non metals with a low electron negativity will lose electrons when reacting with another non metal that has a higher electron negativity. When the non metal forms an ion it is necessary to look at the electron structure to determine how many electrons the element can lose to gain stability.

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6 0
3 years ago
As temperature increases, density increases.<br><br> True or False
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