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Tema [17]
3 years ago
15

How are monosaccharides and polysaccharides structurally different? *NO PLAGIARISM PLS*

Chemistry
2 answers:
Brut [27]3 years ago
8 0
  • C personalities are organized and controlled. They take their time completing tasks accurately, like to research things thoroughly, and are always prepared.  Because they put so much time and effort into everything they do, they hate to be criticized. They are perfectionists at heart and take even the smallest complaint personally.  This makes them extremely difficult to confront. The best way to approach it is to focus on parts of the problem largely out of the C personality’s control or aspects of the problem they might not have considered.  Make it clear that you appreciate all of their efforts and that you’re bringing something to their attention because it’s a variable that they need to have in the equation in the future. No matter what, don’t frame the conflict as a personal attack. If you do, the C personality will likely become resentful and go out of their way to avoid future interactions with you.!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!<u>!</u><u>!</u><u>!</u><u>!</u><u>!</u><u>!</u><u>!</u><u>!</u><u>!</u><u>!</u>
Mariana [72]3 years ago
3 0

Explanation:

Monosaccharides are simple carbohydrates that cannot be further hydrolyzed to simpler carbohydrates. They contain between three and six carbon atoms per molecule.

Polysaccharides  are complex carbohydrates . They are condensation polymers derived from very long chains of monosaccharide units.

Structurally, polysaccharides are made up of repeating units of monosaccharides.

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3 years ago
g Consider a uniport system where a carrier protein transports an uncharged substance A across a cell membrane. Suppose that at
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Answer :  The ratio of the concentration of substance A inside the cell to the concentration outside is, 296.2

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln Q\\\\\Delta G^o=-RT\times \ln (\frac{[A]_{inside}}{[A]_{outside}})

where,

\Delta G^o = standard Gibbs free energy  = -14.1 kJ/mol

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T = temperature = 25^oC=273+25=298K

Q  = reaction quotient

[A]_{inside} = concentration inside the cell

[A]_{outside} = concentration outside the cell

Now put all the given values in the above formula, we get:

-14.1\times 10^3J/mol =-(8.314J/K.mol)\times (298K)\times \ln (\frac{[A]_{inside}}{[A]_{outside}})

\frac{[A]_{inside}}{[A]_{outside}}=296.2

Thus, the ratio of the concentration of substance A inside the cell to the concentration outside is, 296.2

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3 years ago
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