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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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3 years ago
If a4 = 625, what does (a) equal?
Reika [66]

I'm assuming you meant to write a^4 = 625.

If that is the case, then note how 625 = 25^2, and how a^4 is the same as (a^2)^2

So we go from this

a^4 = 625

to this

(a^2)^2 = 25^2

Apply the square root to both sides and you'll end up with: a^2 = 25

From here, apply the square root again to end up with the final answer: a = 5 or a = -5

As a check:

a^4 = (-5)^4 = (-5)*(-5)*(-5)*(-5) = 25*25 = 625

a^4 = (5)^4 = (5)*(5)*(5)*(5) = 25*25 = 625

Both values of 'a' work out

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Answer:

Kawaii, the problem is the comparison with the calculated time for Jillian and the actual time for Jillian to run the 7 miles.

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