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3 years ago

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

Chemistry

1 answer:

mario62 [17]3 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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The three types of nuclear radiation in increasing order of penetrating power are ____.

cricket20 [7]

Answer:

The three types of nuclear radiation in increasing order of penetrating power are ____.

alpha, beta, gamma

Explanation:

Alpha Ray: This has the least penetrating power because the particles produced during decay are large in quantity but they have a very low energy which dosent alow them to move far through space because the are ususally blocked.

Beta ray : they have more penetrating power than alpha rays because they a bit higher in energy and quantity {size}

Gamma Ray; this has the most penetrating power. they are highly powerful waves but do little at ionization of other atoms or molecules. it penetrates through molecules very easily due to its size and energy.

5 0

3 years ago

Read 2 more answers

Calculate the second volumes. <br> 7.03 Liters at 31 C and 111 Torr to STP

DIA [1.3K]

The second volume : V₂= 0.922 L

<h3> Further explanation </h3><h3>Given </h3>

7.03 Liters at 31 C and 111 Torr

Required

The second volume

Solution

T₁ = 31 + 273 = 304 K

P₁ = 111 torr = 0,146 atm

V₁ = 7.03 L

At STP :

P₂ = 1 atm

T₂ = 273 K

Use combine gas law :

P₁V₁/T₁ = P₂V₂/T₂

Input the value :

0.146 x 7.03 / 304 = 1 x V₂/273

V₂= 0.922 L

4 0

3 years ago

If you feed 100 kg of N2 gas and 100 kg of H2 gas into a

torisob [31]

Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of $N_2$ = 100 kg = 100000 g

Mass of $H_2$ = 100 kg = 100000 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $H_2$ = 2 g/mole

Molar mass of $NH_3$ = 17 g/mole

First we have to calculate the moles of $N_2$ and $H_2$ .

$\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles$

$\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2+3H_2\rightarrow 2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $N_2$ react with 3 mole of $H_2$

So, 3571.43 moles of $N_2$ react with $3571.43\times 3=10714.29$ moles of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $N_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NH_3$

From the reaction, we conclude that

As, 1 mole of $N_2$ react to give 2 mole of $NH_3$

So, 3571.43 moles of $N_2$ react to give $3571.43\times 2=7142.86$ moles of $NH_3$