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valkas [14]
3 years ago
7

The labels on a shipment of the five chemicals A –E below were washed off by accident when the delivery truck went through a tru

ckwash with the doors open. A clever student took infrared spectra of each of the resulting unknowns. Unknown #1 had a strong, broad IR absorption at 3400–3600 cm–1, and no absorptions between 1600 and 2900 cm–1. Unknown #2 had a strong absorption at 1680 cm–1 and no absorptions above 3100 cm–1. What are the structures of unknowns 1 and 2?
Mathematics
1 answer:
Tanya [424]3 years ago
8 0

Answer:

Unknown 1: -OH group at 3400-3600

Unknown 2: Ketone and alkene group

Step-by-step explanation:

Unknown 1:

This compound has a strong IR absorption at 3400 - 3600cm-¹; this indicates that it is an alcoholic group (OH)

No absorption between 1600 and 2900cm-¹: this rule out the presence of ketone (C = O)

Hence the unknown compound is OH group

Unknown 2:

This has strong IR absorption at 1680cm-¹: this indicates the presence of (C=O) group.

No absorptions above 3100cm-¹: this indicates the presence of C = C - H

So therefore, unknown 2 is most likely to be Ketone and alkene group

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0.071,1.928

Step-by-step explanation:

                                                Downtown Store   North Mall Store

Sample size   n                             25                        20

Sample mean \bar{x}                         $9                        $8

Sample standard deviation  s       $2                        $1

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\bar{x_1}=9\\ \bar{x_2}=8

s_1=2\\s_2=1

x_1-x_2=9-8=1

Standard error of difference of means = \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}

Standard error of difference of means = \sqrt{\frac{2^2}{25}+\frac{1^2}{20}}

Standard error of difference of means = 0.458

Degree of freedom = \frac{\sqrt{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1}+\frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}

Degree of freedom = \frac{\sqrt{(\frac{2^2}{25}+\frac{1^2}{20}})^2}{\frac{(\frac{2^2}{25})^2}{25-1}+\frac{(\frac{1^2}{20})^2}{20-1}}

Degree of freedom =36

So, z value at 95% confidence interval and 36 degree of freedom = 2.0280

Confidence interval = (x_1-x_2)-z \times SE(x_1-x_2),(x_1-x_2)+z \times SE(x_1-x_2)

Confidence interval = 1-(2.0280)\times 0.458,1+(2.0280)\times 0.458

Confidence interval = 0.071,1.928

Hence Option A is true

Confidence interval is  0.071,1.928

4 0
3 years ago
What is 2/3 - 1/2 <br> &amp; <br> 2/5 - 1/8
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11/40
5 0
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