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Annette [7]
3 years ago
13

Which metal atoms can form ionic bonds by losing electrons from both the outermost and next to outermost principal energy levels

?
Chemistry
1 answer:
RoseWind [281]3 years ago
5 0

Full question options;

(Fe, Pb, Mg, or Ca)

Answer:

Iron - Fe

Explanation:

We understand tht metals pretty much form bonds by losing their valence (outermost electrons). But this question specifically asks for metals that lose beyond their outermost electrons; next to outermost principal energy levels.

Pb, Mg, and Ca only lose their outermost electrons to form the following ions;

Pb2+, Mg2+, and Ca2+.

This is because their ions have achieved a stable octet configuration - the dreamland of atoms where they are satisfied and don't need to go into reactions again.

Iron on the other hand has the following electronic configurations;

Fe:  [Ar]4s2 3d6

Fe2+:  [Ar]4s0 3d6

Fe3+:  [Ar]4s0 3d5

This means ion can lose both the ooutermost electrons (4s) and next to outermost principal energy levels (3d). So correct option is Iron.

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What is the conclusion of an experiment
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Part 1. A chemist reacted 18.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equat
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Answer:

Part 1

The mass of the NaCl that reacted with F₂ at 290.K and 1.5 atm is approximately 132.6 gams

Part 2

The mass of NaCl that can react with the same volume of gas at STP is approximately 93.77 grams

Explanation:

Part 1

The volume of F₂ gas in the reaction, V = 18.0 liters

The ideal gas equation is P·V = n·R·T

∴ n = P·V/(R·T)

The pressure, P = 1.5 atm

The temperature, T = 290 K

The universal gas constant, R = 0.0820573 L·atm/(mol·K)

∴ n = 1.5×18/(0.0820573 × 290) ≈ 1.134615

The number of moles of F₂ in the reaction n ≈ 1.134615 moles

The chemical reaction is given as follows;

F₂ + 2NaCl → Cl₂ + 2NaF

1 mole of F₂ reacts with 2 moles of NaCl

Therefore;

1.134615 moles of F₂ reacted with 2 × 1.134615 moles ≈ 2.26923 moles of NaCl

1 mole of NaCl = The molar mass of NaCl, MM = 58.44 g/mol

The mass, of 2.26923 moles of NaCl, m = Number of moles × MM

∴ m ≈ 2.26923 moles × 58.44 g/mol ≈ 132.6 grams

The mass of the NaCl ≈ 132.6 gams

Part 2

The volume occupied by 1 mole of all gases at STP = 22.4 l/mole

Therefore, the number of moles of F₂ in 18.0 L of F₂ = 18.0 L/(22.4 L/mole) ≈ 0.804 moles

Therefore;

The number of moles of NaCl, in the reaction n = 2 × The number of moles of F₂ ≈ 2×0.804 moles = 1.608 moles

The number of moles of NaCl, in the reaction n ≈ 1.608 moles

The mass of NaCl in the reaction, m = n × MM

∴ m ≈ 1.608 moles × 58.44 g/mol ≈ 93.97 grams

The mass of NaCl that can react with the same volume of gas at STP ≈ 93.77 grams

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