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kifflom [539]
3 years ago
14

Predict whether each of the following bonds is ionic, polar, covalent, or nonpolar covalent:

Chemistry
2 answers:
uysha [10]3 years ago
8 0

Predict whether each of the following bonds is no polar covalent, polar covalent, or ionic

Bond. Electro negativity Type of bond

Si- O

k-Cl

I-I

C-H

Gekata [30.6K]3 years ago
7 0

Explanation:

It is known that in order to determine a compound to be ionic, polar, covalent or non-polar covalent it is necessary to determine the electronegativity difference of the combining atoms.

So, when electronegativity difference is from 0.0 to 0.4 then bond formed between the two atoms is non-polar covalent in nature.

When electronegativity difference is greater than 0.4 and less than 1.8 then bond between the two atoms is a polar covalent bond.

When electronegativity difference is 1.8 or greater than the bond formed is ionic in nature.

Electronegativity difference of the given molecules is as follows.

Si-O = (1.90 - 3.44) = 1.54

K-Cl = (0.82 - 3.16) = 2.34

S-F = (2.58 - 3.98) = 1.4

P-Br = (2.19 - 2.96) = 0.77

Li-O = (0.98 - 3.44) = 2.46

N-P = (3.04 - 2.19) = 0.85

Therefore, given compounds are classified as follows.

Si-O, S-F, P-Br, and N-P all have a polar covalent bond. Whereas K-Cl and Li-O are ionic in nature.

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Natali5045456 [20]

There are 6.33 × 10²⁵ hydrogen atoms in this solution in total.

<h3>Explanation</h3>
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2.10 × 10²⁵ water molecules and 7.10 × 10²⁴ ammonia molecules will contain

2 \times 2.10 \times 10^{25} + 3 \times 7.10 \times 10^{24} = 6.33 \times 10^{25} hydrogen atoms in total.

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How does the brønsted-Lowry theory define acids and bases?
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They define acids as proton donors, and bases as proton acceptors

If you were to have:

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You can see that the nitric acid (HNO3) gave a hydrogen ion which has 1 proton, 0 neutrons and 0 electrons to the water so we just say that it gave a proton.

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8 0
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Answer:

The Structure of "B" is alkene.

Explanation:

The compound "A" having R- configuration  and undergoes Hofmann elimination to form an alkene.

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