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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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A customer experiences worsening side effects in response to a prescription. What do you suggest to them?
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Arrange the following atoms in order of increasing first ionization energy: He, Be, Se, Ne
Alex17521 [72]

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Be (899 kj/mol) , Se (940.9 kj/mol), Ne(2081 kj/mol), He (2370 kj/mol),

Explanation:

For noble gases as they have complete octet so they require high amount of energy to remove the electron.

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As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.

Trend along group:

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Can someone show their work on how to do this: Calculate the number of moles of Al2O3 present in 23.87g. (Molar mass of Al2O3= 1
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The reaction of ethane gas (C2H6) with chlorine gas produces C2H5Cl as its main product (along with HCl). In addition, the react
Kazeer [188]

Answer:

The percent yield of  chloro-ethane in the reaction is 82.98%.

Explanation:

C_2H_6+Cl_2\rightarrow C_2H_5Cl+HCl

Moles of ethane = \frac{300.0 g}{30 g/mol}=10 mol

Moles of chlorine gases =\frac{650.0 g}{71 .0 g/mol}=9.1549 mol

As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.

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According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.

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\frac{1}{1}\times 9.1549 mol=9.1549 mol of chloro-ethane

Mass of 9.1549 moles of chloro-ethane:

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Given experimental yield of chloro-ethane: 490.0 g

\% Yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

\%Yield (C_2H_5Cl)=\frac{490.0 g}{590.4910 g}\times 100=82.98\%

The percent yield of  chloro-ethane in the reaction is 82.98%.

6 0
3 years ago
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