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lubasha [3.4K]
3 years ago
9

Robert has pure samples of both D-ribose and D-arabinose, but he forgot to label them. He only has some nitric acid, the reagent

s for a Wohl degradation, and a polarimeter (an instrument for measuring optical activity) in his lab. In approximately 50 words or less, explain how Robert can determine which sample is which without using any additional resources. If it is not possible, clearly state so and explain why it is not possible in approximately 50 words or less.
Chemistry
1 answer:
Sauron [17]3 years ago
4 0

Answer:

Following are the responses to the given question:

Explanation:

Since HN03 is an oxidation substance D-ribose u.ith oxidized to form in rubric acid Ribose is chiral, but rubric acid is achiral because of its symmetry mirror level, Hence no infrared roster in the sample holder is observed.

Please find the attached file.

D-Arabinose, on either hand, gives optical aldaric acid with such a net optical rotation observed inside the polarimeter for diagnosis with HN03.

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IgorLugansk [536]

<u>Answer:</u> The final temperature of the mixture is 51.49°C

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Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]       ......(1)

where,

q = heat absorbed or released

m_1 = mass of water at high temperature = 140 g     (Density of water = 1.00 g/mL)

m_2 = mass of water at low temperature = 230 g

T_{final} = final temperature = ?°C

T_1 = initial temperature of water at high temperature = 95.00°C

T_2 = initial temperature of water at low temperature = 25.00°C

c = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

140\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]

T_{final}=51.49^oC

Hence, the final temperature of the mixture is 51.49°C

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How should the baking of a pizza be categorized
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4 years ago
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Explanation:

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