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4 years ago

A(n) _______________ can be formed by linking together several monosaccharides via glycosidic bonds.

Chemistry

1 answer:

mario62 [17]4 years ago

7 0

Answer:

A polysaccharide (n) can be formed by linking several monosaccharides through glycosidic linkages.

Explanation:

Polysaccharides are carbohydrates or complex carbohydrates, where monosaccharides join with glucosidic bonds to form a more complex structure that would be the polysaccharide.

An example of a polysaccharide is starch, or glycogen.

Starch is found in many foods such as potatoes or rice, and glycogen is a form of energy reserve of our organism housed in muscles and liver to fulfill locomotion, physical activity, and other activities that consist of glycolysis.

Polysaccharides are degraded in our body by different stages, and several enzymes unlike monosoccharides or disaccharides, since they have more unions and a more complex structure to disarm in our body and thus assimilate it.

Polysaccharides are also part of animal structures, such as insect shells or nutritional sources, among others.

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Suppose of barium acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of barium cati

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Explanation:

Let us assume that the given data is as follows.

mass of barium acetate = 2.19 g

volume = 150 ml = 0.150 L (as 1 L = 1000 ml)

concentration of the aqueous solution = 0.10 M

Therefore, the reaction equation will be as follows.

$Ba(C_{2}H_{3}O_{2})_{2} \rightarrow Ba^{2+} + 2C_{2}H_{3}O^{-}_{2}$

Hence, moles of $C_{2}H_{3}O^{-}_{2}$ = $2 \times Ba(C_{2}H_{3}O_{2})_{2}$ .......... (1)

As, No. of moles = $\frac{mass}{\text{molar mass}}$

Hence, moles of $Ba(C_{2}H_{3}O_{2})_{2}$ will be calculated as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{2.19 g}{255.415 g/mol}$ (molar mass of $Ba(C_{2}H_{3}O_{2})_{2}$ is 255.415 g/mol)

= $8.57 \times 10^{-3}$

Moles of $C_{2}H_{3}O^{-}_{2}$ = $2 \times 8.57 \times 10^{-3}$

= 0.01715 mol

Hence, final molarity will be as follows.

Molarity = $\frac{\text{no. of moles}}{volume}$

= $\frac{0.01715 mol}{0.150 L}$

= 0.114 M

Thus, we can conclude that final molarity of barium cation in the solution is 0.114 M.

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A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

aniked [119]

<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

Putting values in above equation, we get:

$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g$