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r-ruslan [8.4K]
2 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]2 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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Calculate the equilibrium constant for the reaction between fe2+(aq) and zn(s) under standard conditions at 25∘c.
Hatshy [7]
Following reaction occurs in the given electrochemical system:
Fe^{2+} + Zn → Fe + Zn^{2+}
Thus, under standard conditions
E(0) = E(0) Fe2+/Fe - E(0) Zn2+/Zn
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E^{0}Zn2+/Zn = standard reduction potential of Zn2+/Zn = -0.763 v

E(0) = 0.323 v
now, we know that, ΔG(0) =-nFE(0) ............... (1)
Also, ΔG^{0} = -RTln(K) ............ (2)

On equating and rearranging equation 1 and 2, we get
K = exp( \frac{nFE(0)}{RT} )= exp (\frac{2X96500X0.323}{8.314X298}) = 8.46 x 10^{10}

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2 years ago
Please help me asap ​
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3 years ago
The rate constant for this first‑order reaction is 0.550 s−10.550 s−1 at 400 ∘C.400 ∘C. A⟶products A⟶products How long, in secon
Alisiya [41]

Answer:

t=2.08s

Explanation:

Hello,

In this case, for first order reactions, we can use the following integrated rate law:

ln(\frac{[A]}{[A]_0} )=kt

Thus, we compute the time as shown below:

t=-\frac{ln(\frac{[A]}{[A]_0} )}{k}=- \frac{ln(\frac{0.220M}{0.690M} )}{0.55s^{-1}} \\\\t=-\frac{-1.14}{0.550s^{-1}}\\ \\t=2.08s

Best regards.

5 0
3 years ago
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