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Andrei [34K]
3 years ago
6

g Which ONE of the following pure substances will exhibit hydrogen bonding? A) methyl fluoride, FCH3 B) dimethyl ether, CH3C–O–C

H3 C) formaldehyde, H2C=O D) trimethylamine, N(CH3)3 E) hydrazine, H2N-NH2
Chemistry
1 answer:
luda_lava [24]3 years ago
8 0

Answer:

C) formaldehyde, H2C=O.

Explanation:

Hello,

In this case, given that the hydrogen bondings are known as partial intermolecular interactions between a lone pair on an electron rich donor atom, particularly oxygen, and the antibonding molecular orbital of a bond between hydrogen and a more electronegative atom or group. Thus, among the options, C) formaldehyde, H2C=O, will exhibit hydrogen bonding since the lone pair of electrons of the oxygen at the carbonyl group, are able to interact with hydrogen (in the form of water).

Best regards.

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Compound Y has the elements C, H, N and O. It's exact mass is 91.0425. Answer the following questions. Isotope Exact Mass 1H 1.0
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Answer:

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Explanation:

The rule of 13 states that the formula of a compound is a multiple n of 13 (the molar mass of CH) plus a remainder r.

MF = CₙHₙ₊ᵣ

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91/13 =7r0

The formula can't be C₇H₇ because a hydrocarbon must have an even number of H atoms,

The odd mass and the odd number of H atoms make it reasonable to add an N atom and subtract CH₂ (CH₂ = 14):

C₇H₇ + N - CH₂ = C₆H₅N

Check:

6C = 6 × 12.000 = 72.000 u

5H = 5 ×   1.008 =   5.040

1N =  1 × 14.003 =  <u>14.003    </u>

             TOTAL =   91.043 u

This is excellent agreement with the observed mass of 91.0425 u.

There are six  C atoms (C₆)

There are five H atoms (H₅)

There is    one N atom   (N)

There are no   O atoms.

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