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maria [59]
3 years ago
12

The formula for a compound of Li₊ ions and Br_ ions is written LiBr. Why can't it be written Li₂Br? Why isn't it written BrLi?

Chemistry
2 answers:
Ahat [919]3 years ago
7 0

Answer: Group 1 elements have an oxidation charge of +1. They are alkali metals and Li is one of them. For Br it is a halogen and in group 7 so it has an oxidation of -1. If we do the criss cross rule it will ha Li+¹ Br-¹. Put each charges interchangeably on each elements. So the charge of Li will be the subscript of Br and vice versa. Since both have the same charge they will cancel out on the final equation of the compound.

Explanation:

bogdanovich [222]3 years ago
5 0

Answer:

.

Explanation:

Lithium is in the first column of the periodic table, so it will have 1 valence electron.

Bromine is in the seventh column of the periodic table, so it will have seven valence electrons.

They must combine in a way to reach 8.

When combining elements to form compounds, the "crisscross method" is used. Above Li would be a charge of +1, and above Br would be a charge of -1.

Cross the 1 from the top of Li to the bottom of Br, and so there is 1 Br.

Cross the 1 from the top of Br to the bottom of Li, and so there is 1 Li.

It is not written BrLi because chemists decided to order them the other way. Technically speaking, it isn't wrong, but the positive charge is normally put on the left and the negative charge is normally put on the right.

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hram777 [196]

Explanation:

Teniendo en cuenta los numeros de oxidacion negativos de cada uno

P:fosforo(-3)

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Y el H: hidrogeno con una valencia

positiva de +1

los compuesto que se formaran son los siguientes

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6 0
2 years ago
Consider the reaction: I2(g)+Cl2(g)⇌2ICl(g) Kp= 81.9 at 25 ∘C. Calculate ΔGrxn for the reaction at 25 ∘C under each condition: -
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Answer:

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Part 2: - 1.137 x 10⁴ J/mol.

Explanation:

Part 1: At standard conditions:

At standard conditions Kp= 81.9.

∵ ΔGrxn = -RTlnKp

∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(81.9)) = - 1.091 x 10⁴ J/mol.

Part 2: PICl = 2.63 atm; PI₂ = 0.324 atm; PCl₂ = 0.217 atm.

For the reaction:

I₂(g) + Cl₂(g) ⇌ 2ICl(g).

Kp = (PICl)²/(PI₂)(PCl₂) = (2.63 atm)²/(0.324 atm)(0.217 atm) = 98.38.

∵ ΔGrxn = -RTlnKp

∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(98.38)) = - 1.137 x 10⁴ J/mol.

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