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Colt1911 [192]
3 years ago
13

Make a model for the the ionic bonding of Al and Cl.

Chemistry
1 answer:
Bezzdna [24]3 years ago
7 0

Answer:

Cl_3Al

Explanation:

* = electron

Cl (17) * * * * * * *

Al (13) * * *

If one of aluminum's electrons were added to chlorine to make a full shell, still have 2 left over. So you need another:

Cl * * * * * * * * (full)

Cl * * * * * * *

Al * *

Still have another left over. Aluminum needs to lose another one of its valence electrons to be able to have a full shell. So you take another chlorine:

Cl * * * * * * * * (full)

Cl * * * * * * * * (full)

Cl * * * * * * *

Al *

You will eventuate with 3Cl and 1Al, or Cl_3Al. Modeled above:

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in an experiment 3.425g of lead oxide was reduced to form 3.105g of lead the empirical formula of the lead oxide is?​
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Answer:

Pb3O4

Explanation:

According to this question, 3.425g of lead oxide was reduced to form 3.105g of lead in an experiment. Since lead oxide contains both lead (Pb) and oxygen (O) element,

Mass of lead oxide = 3.425g

Mass of lead = 3.105g

Mass of oxygen = (3.425g - 3.105g) = 0.320g

Next, we convert each mass value to mole by dividing by respective molar mass

Pb = 3.105g ÷ 207.2 = 0.0149mol

O = 0.320g ÷ 16 = 0.02mol

Next, we divide each mole value by the smallest (0.0149)

Pb = 0.0149mol ÷ 0.0149mol = 1

O = 0.02mol ÷ 0.0149mol = 1.342

Multiply each ratio value by 3 to get:

Pb = 1 × 3 = 3

O = 1.342 × 3 = 4.026

The whole number ratio, approximately, of Pb and O is 3:4, hence, their empirical formula is Pb3O4.

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