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r-ruslan [8.4K]
3 years ago
6

The difference in interchain stability between the polysaccharides glycogen and cellulose is due to: Group of answer choices the

incorporation of complex ions in the three dimensional structures of both polysaccharides. both the different glycosidic linkages of the molecules and the different hydrogen bonding partners of the individual chains. the different hydrogen bonding partners of the individual chains. None of the answers is correct the different glycosidic linkages of the molecules.
Chemistry
1 answer:
denis-greek [22]3 years ago
7 0

Answer: both the different glycosidic linkages of the molecules and the different hydrogen bonding partners of the individual chains.

Explanation:

Glycogen is a polysaccharide of glucose which is a form of energy storage in fungi, bacteria and animals. Glycogen is primarily stored in the liver cells and skeletal muscle.

The difference in interchain stability between the polysaccharides glycogen and cellulose is due to the different glycosidic linkages of the molecules and the different hydrogen bonding partners of the individual chains.

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Is boiling water exothermic or endothermic ? What about condensing?
Ugo [173]

Answer:

exothermic

Explanation:

boiling water releases heat and is therfor exothermic

condensing is the reverse reaction and is endothermic

8 0
4 years ago
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You are given two aqueous solutions with different ionic solutes (Solution A and Solution B). What if you are told that Solution
klio [65]

Answer:

Yes, it is possible. Let us consider an example of two solutions, that is, solution A having 20 percent mass RbCl (rubidium chloride) and solution B is having 15 percent by mass NaCl or sodium chloride.  

It is found that solution A is having more concentration in comparison to solution B in terms of mass percent. The formula for mass percent is,  

% by mass = mass of solute/mass of solution * 100

Now the formula for molality is,  

Molality = weight of solute/molecular weight of solute * 1000/ weight of solvent in grams

Now molality of solution A is,  

m = 20/121 * 1000/80 (molecular weight of RbCl is 121 grams per mole)

m = 2.07

Now the molality of solution B is,  

m = 15/58.5 * 1000/85

m = 3.02

Therefore, in terms of molality, the solution B is having greater concentration (3.02) in comparison to solution A (2.07).

5 0
3 years ago
Determine the genotypes of offspring 1-8.
8_murik_8 [283]
Do you have a picture
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3 years ago
If Jonathan went skateboarding from 4:00 PM to 4:30 PM and traveled 450 meters, what was his average speed?
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8 0
3 years ago
Calculate the pH of a solution that is 2.00 M HF, 1.00 M NaOH, and 0.472 M NaF. (Ka=7,2×104)
eduard

Explanation:

The given data is as follows.

Concentration of acid, [HF] = 2 M

Concentration of salt, [NaF] = 0.472 M

Concentration of base, [NaOH] = 1.00 M

K_{a} = 7.2 \times 10^{-4}

So, NaOH being a base will react with the acid that is, HF. And, according to Henderson-Hasselbalch equation we have he following.

             pH = pK_{a} + log\frac{[salt]}{[acid]}

Also, we known that pK_{a} = -log K_{a}

so,                     pK_{a} = -log 7.2 \times 10^{-4}

                         pK_{a} = 3.1426

Hence,    pH = 3.1426 + log\frac{0.472 M}{2 M}

               pH = 2.5155

Thus, we can conclude that pH of the solution is 2.5155.

8 0
3 years ago
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