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meriva
1 year ago
6

An element forms an oxide, E₂O₃, and a fluoride, EF₃.(a) Of which two groups might E be a member?

Chemistry
1 answer:
Fittoniya [83]1 year ago
8 0

a) The E might belong to group 13.

As the formula of a chemical compound is derived by the cross multiplication of the valency of the atoms. As formula of the given oxide is  and valency of O atom is -2, therefore valency of element E must be +3 in order to obtain E2O3.

Also, in EF3, the valency of E will be +3 because there are three atoms of fluorine who has an individual valency of -1. Thus, e will have the valency of +3.

The Group 13 is the boron group which has the following elements:

  • Boron
  • Aluminium
  • Gallium
  • Indium
  • Thallium

All these elements have the valency of +3.

To know more about Valency, refer to this link:

brainly.com/question/12717954

#SPJ4

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2 years ago
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How much energy is evolved during the formation of 197 g of Fe, according to the reaction below?
hoa [83]

Answer:

ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'

Explanation:

Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj

197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)

From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...

3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).

______

NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g.  Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.  

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2 years ago
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