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s344n2d4d5 [400]
2 years ago
8

5. Glycogen and starch both give a (+) test with I2-KI and a (-) test with Benedict's. If you hydrolyzed these two polysaccharid

es and tested the solutions with the same two reagents what results would you get? Explain these results.
Chemistry
1 answer:
SVETLANKA909090 [29]2 years ago
3 0

Answer:

There will be no reaction between the product of hydrolysis and I2-KI (-ve). When the product of hydrolysis is tested with Benedict reagent, a brick-red precipitate is observed.

Explanation:

Benedict's reagent as a chemical reagent is a mixture of sodium carbonate, sodium citrate and copper(II) sulfate pentahydrate. It is often used in place of Fehling's solution to detect the presence of reducing sugars. A positive test with Benedict's reagent is shown by a color change from clear blue to a brick-red precipitate.

Glycogen and starch are both complex structures containing repeating units of glucose(a reducing sugar). The polysaccharides have non redusing ends and so cannot react with benedict reagent.

When they are hydrolysed, glucose which is a reducing sugar can then test positive with Benedict reagent.

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4 0
2 years ago
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1. Which liquid sample is a pure substance?
IRISSAK [1]

The whole Activity , poem and paragraph is missing in the question.

Answer:

(1) Liquid A

(2) Solid A

Explanation:

Using this part of the given poem

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Solid A is melting completely so Solid A is a pure substance.

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Sulfurous acid is a diprotic acid with the following acid-ionization constants: Ka1 = 1.4x10−2, Ka2 = 6.5x10−8 If you have a 1.0
REY [17]

Answer:

pH = 7.1581

Explanation:

The equilibrium of NaHSO₃ with Na₂SO₃ is:

HSO₃⁻ ⇄ SO₃²⁻ + H⁺

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HSO₃⁺ reacts with NaOH, thus:

HSO₃⁻ + NaOH → SO₃²⁻ + H₂O + Na⁺

As the buffer is of 1.0L, initial moles of HSO₃⁻ and SO₃²⁻ are:

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SO₃²⁻: 0.139 moles

Based on the reaction of NaOH, moles added of NaOH are subtracting moles of HSO₃⁻ and producing SO₃²⁻. The moles added are:

0.0500L ₓ (1mol /L): 0.050 moles of NaOH.

Thus, final moles of both compounds are:

HSO₃⁻: 0.252 moles - 0.050 moles = 0.202 moles

SO₃²⁻: 0.139 moles + 0.050 moles = 0.189 moles

Using H-H equation for the HSO₃⁻ // SO₃²⁻ buffer:

pH = pka + log [SO₃²⁻] / [HSO₃⁻]

Where pKa is - log Ka = 7.187

Replacing:

pH = 7.187 + log [0.189] / [0.202]

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4 0
2 years ago
Which one of the equations below is an exothermic reaction?
Yuri [45]

Answer:

B) CaO(s) + H2O(l) --> Ca(OH)2(aq)

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This is the only reaction with a negative enthalpy value. Exothermic reactions have a negative enthalpy.

3 0
2 years ago
An evaporation–crystallization process of the type described in Example 4.5-2 is used to obtain solid potassium sulfate from an
timofeeve [1]

Answer:

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The rate of supplying fresh feed to obtain the production rate is:

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7 0
3 years ago
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