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VikaD [51]
3 years ago
6

What characteristic frequencies in the infrared spectrum of your sodium borohydride reduction product will you look for to deter

mine whether the carbonyl group (in ethyl vanillin) has been converted to an alcohol functional group? That is, if the transformation has occurred, which characteristic frequencies should be present in the infrared spectrum, and which should be absent? How would the infrared spectrum for the sodium borohydride reduction product be affected if the product were still wet with water? Explain your reasoning.
Chemistry
1 answer:
kkurt [141]3 years ago
5 0

Answer:

A)The characteristic frequency to look out for is  1720-1740 cm-1 (for C=O) for which will disappear in the end product but initially present in the reactant.

B)Characteristic frequency  present in the infrared spectrum will be at a peak of  3300-3400 cm-1 which will be due to O-H stretch.

C)If the product is wet with water there will be no change in the infrared spectrum

Explanation:

The characteristic frequency to look out for is  1720-1740 cm-1 (for C=O) for which will disappear in the end product but initially present in the reactant.

Characteristic frequency  present in the infrared spectrum will be at a peak of 3300-3400 cm-1 which will be due to O-H stretch.

If the product is wet with water there will be no change in the infrared spectrum

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acid, HNO3

base, KOH

salt ,KNO3

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Which could be the pH of an aqueous solution with a H3O+ ion concentration that is less than its OH- ion concentration?
anzhelika [568]

Answer:

Explanation:

You can think of pH as "parts Hydrogen ion," but remember that the pH scale is "backwards." The pH scale ranges from 0 to 14, with zero being the most acidic (highest concentration of H+) and 14 being the most basic.

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What is produced when an acid and an alkali react?
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Read 2 more answers
The masses of carbon and hydrogen in samples of four pure hydrocarbons are given above. The hydrocarbon in which sample has the
enot [183]

Answer:

Sample B

Explanation:

In this case, we need to determine the empirical formula for each sample. The one that match the formula of the propene would be the sample.

Let's do Sample A:

C: 60 g;       H: 12 g

1. Calculate moles:

We need the atomic weights of carbon (12 g/mol) and hydrogen (1 g/mol):

C: 60 / 12 = 5

H: 12 / 1 = 12

2. Determine number of atoms in the formula

In this case, we just divide the lowest moles obtained in the previous part, by all the moles:

C: 5 / 5 = 1

H: 12 / 5 = 2.4    or rounded to two

3. Write the empirical formula:

Now, the prior results, represent the number of atoms in the empirical formula for each element, so, we put them with the symbol and the atoms as subscripted:

C₁H₂ = CH₂

Therefore, sample A is not the same as propene.

Sample B:

C: 72 g    H: 12 g

Following the same steps, let's determine the empirical formula for this sample

C: 72 / 12 = 6 ---> 6 / 6 = 1

H: 12 / 1 = 12 ----> 12 / 6 = 2

EF: CH₂

Sample C:

C: 84 g    H: 10 g

C: 84 / 12 = 7 ----> 7 / 7 = 1

H: 10 / 1 = 10    ----> 10 / 7 = 1.4 or just 1

EF: CH

Sample D

C: 90 g      H: 10 g

C: 90 / 12 = 7.5     -----> 7.5 / 7.5 = 1

H: 10 / 1 = 10  -------> 10 / 7.5 = 1.33 or just 1

EF: CH

Neither compound has the same empirical formula as C3H6, but C3H6 is a molecular formula, so, if we just simplify the formula we have:

C3H6  -----> CH₂

Therefore, sample B is the one that match completely. Sample B would be the one.

Hope this helps

8 0
3 years ago
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