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3 years ago

Is Cellulose ketohexose or aldohexose? How does Cellulose give result in Seliwanoff’s test (any color change)?

Chemistry

1 answer:

Mamont248 [21]3 years ago

4 0

Answer:

Cellulose is an aldohexose

Explanation:

Cellulose is an aldohexose , because , it has an aldehyde group present at one of the end .

Seliwanoff’s test

It is the test to differentiate between a ketose and an aldose .

The test is as follows -

If the sample is heated and it it readily gets dehydrated , then the sample is a ketose .

And , the reaction of Seliwanoff’s reagent and ketose , give a red color , which is the positive sign for the test .

But ,

the reaction of Seliwanoff’s reagent and aldoses , give a very light pink color .

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Question:

Part D) An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in joules per kelvin.

Part E) An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part F) An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part G) Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the temperatures of the objects do not change appreciably in the process. Express your answer numerically in joules per kelvin.

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Explanation:

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Part D)

Given:

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Entropy change will be:

ΔS = $\frac{25*1000}{293}$

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Part E)

Given:

T = 500K

Q = -25.0 kJ

Entropy change will be:

ΔS = $\frac{-25*1000}{500}$

= - 50 J/K

Part F)

Given:

T = 400K

Q = 25.0 kJ

Entropy change will be:

ΔS = $\frac{25*1000}{400}$

= 62.5 J/K

Part G:

Given:

T1 = 400K

T2 = 500K

Q = 25.0 kJ

The net entropy change will be:

ΔS = $(\frac{25*1000}{400}) + (\frac{-25*1000}{500}$

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