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VMariaS [17]
3 years ago
13

Lead is malleable, so it can be pounded into flat sheets without breaking. How does the bonding within lead help to explain this

property?
Metallic bonds involve valence electrons that are removed from one atom and given to another, so the pounding helps the electrons move.
Covalent bonds involve valence electrons that are shared between two metal atoms, so the bonds are strong enough to resist the pounding.
Metallic bonds involve many valence electrons shared by many atoms, so the bonds can move around as the metal is pounded.
Covalent bonds involve valence electrons that are removed from one atom and given to another, so the pounding helps the electrons move.
Chemistry
2 answers:
jarptica [38.1K]3 years ago
9 0

The answer is Metallic bonds involve many valence electrons shared by many atoms, so the bonds can move around as the metal is pounded. The metallic bond structure of lead forms a cubic crystal structure and the atoms can roll over one another without breaking the metallic bonds. This is especially because the p orbital electrons of lead can be delocalized and the electrons can be shared with other lead ions in the cubic structure of lead.

Lena [83]3 years ago
7 0

Answer: Option (c) is the correct answer.

Explanation:

In metallic bonding, the valence electrons are involved in bonding with many atoms.

As a result, valence electrons dissociate within their atomic core and form a sea of electrons which helps in binding the positively charged ions.

Thus, we can conclude that the bonding within lead help to explain this property as metallic bonds involve many valence electrons shared by many atoms, so the bonds can move around as the metal is pounded.

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C_2H_6O

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70.44~g~H_2O\frac{1~mol~H_2O}{18~g~H_2O}=~3.9~mol~H_2O

Now, in 1 mol of CO2 we have 1  mol of C and in 1 mol of H_2O we have 1 mol of H. Additionally, if we want to calculate the moles of oxygen we need to <u>calculate the grams of C and O</u> and then do the <u>substraction</u> form the initial amount, so:

2.6~mol~CO_2\frac{1~mol~C}{1~mol~CO_2}\frac{12~g~C}{1~mol~C}=31.25~g~of~C

3.9~mol~H_2O\frac{2~mol~H}{1~mol~H_2O}\frac{1~g~H}{1~mol~H}=7.82~g~of~H

Total~grams=~31.25~+~7.82=39.08~g

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Now we can <u>convert the grams</u> of O to moles, so:

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O=\frac{1.30~mol~O}{1.30~mol~O}=~1

C=\frac{2.6~mol~C}{1.30~mol~O}=~2

H=\frac{7.82~mol~H}{1.30~mol~O}=6

Therefore the formula is C_2H_6O

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