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Bess [88]
4 years ago
5

Which of the following are strongly hydrogen bonded in the liquid phase? A) nitrilesB) esters C) secondary amides D) acid chlori

des
Chemistry
1 answer:
-Dominant- [34]4 years ago
6 0

Answer: Option (C) is the correct answer.

Explanation:

Chemical formula of a secondary amide is R'-CONH-R, where R and R' can be same of different alkyl or aryl groups. Here, the hydrogen atom of amide is attached to more electronegative oxygen atom of the C=O group.

Therefore, the hydrogen atom will be more strongly held by the electronegative oxygen atom. As a result, there will be  strongly hydrogen bonded in the liquid phase of secondary amide.

Whereas chemical formula of nitriles is RCN, ester is RCOOR' and acid chlorides are RCOCl. As no hydrogen bonding occurs in any of these compounds because hydrogen atom is not being attached to an electronegative atom.

Thus, we can conclude that secondary amides are strongly hydrogen bonded in the liquid phase.

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In a tissue that metabolizes glucose via the pentose phosphate pathway, C-1 of glucose would be expected to end up principally i
Mnenie [13.5K]

Answer:

A.

Carbon dioxide

Explanation:

In a tissue that metabolizes glucose via the pentose phosphate pathway, C-1 of glucose would be expected to end up principally in Carbon dioxide

5 0
2 years ago
I NEED HELP ON THIS WORTH 20 POINTS :)
Arturiano [62]

Answer: C. mollusca and arthropoda

                                                                           

Explanation:

4 0
2 years ago
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Which of the following is NOT an example of a molecule <br> A: H2O2<br> B:NCI3<br> C:F<br> D:O3
timama [110]

C. F because it's an atom not a molecule

6 0
3 years ago
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A pure compound is found to be 40.0% carbon by mass, 6.73% hydrogen by mass, and 53.3% oxygen by mass. determine the empirical f
ratelena [41]
Since there is no weight, I would assume that this is a 100g of pure compound.
Okay so I would be changing the percentage to gram to solve for the mole.
So
40.0g C (1 mol C/12.01 g C) = 3.33 mol C
6.73g H (1 mol H/1.01 g H ) = 6.66 mol H
53.3g O (1 mol O/16.00 g O) = 3.33 mol O

With that, two of our moles is 3.33, so we consider that are our 1, as it is also the lowest. Therefore the empirical formula is CH2O
3 0
3 years ago
A 35.66g sample of copper is heated using 600j of energy. if the original temperature of the copper is 85C what is its final tem
ira [324]

This problem is providing the mass, energy, initial temperature and specific heat of a sample of copper that is required to calculate the final temperature.

Thus, we recall the general heat equation:

Q=mC(T_f-T_i)\\

Which has to be solved for the final temperature, T_f as follows:

T_f=T_i+\frac{Q}{mC}

Finally, we plug in the numbers to obtain:

T_f=85\°C+\frac{600J}{35.66g*0.38\frac{J}{g\°C} } \\\\T_f=129.3\°C

However, this result is not given in the choices.

Learn more:

  • brainly.com/question/14383794
8 0
2 years ago
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