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Nesterboy [21]
3 years ago
7

Why do you suppose the structural polysaccharide cellulose does not contain branches? Why do you suppose the structural polysacc

haride cellulose does not contain branches? Branches in the molecule would generate side chains that would almost certainly make it difficult to pack the cellulose molecules into globules, thereby decreasing the flexibility and strength of the globules. Branches in the molecule would generate side chains that would almost certainly make it difficult to pack the cellulose molecules into microfibrils, thereby increasing the rigidity and strength of the microfibrils. Branches in the molecule would generate side chains that would almost certainly make it difficult to pack the cellulose molecules into globules, thereby increasing the flexibility and strength of the globules. Branches in the molecule would generate side chains that would almost certainly make it difficult to pack the cellulose molecules into microfibrils, thereby decreasing the rigidity and strength of the microfibrils.
Chemistry
1 answer:
madam [21]3 years ago
3 0

The correct statement is that branches in the molecule would generate side chains that would almost certainly make it difficult to pack the cellulose molecules into microfibrils, thereby decreasing the rigidity and strength of the microfibrils.

The prime polysaccharide found in the plants, which is responsible for the structural role is cellulose. It is one of the most naturally abundant organic constituents found on Earth. It is an unbranched polymer of glucose residues, which is combined together through beta-1,4 linkages that permit the molecules to produce straight and long chains. The cellulose molecules are linear, rigid rods, which combines laterally into microfibrils.  


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if a solution is saturated, how does the rate of dissolution of a solute compare with it's rate of crystallization?
LenKa [72]

Answer:

The same

Explanation:

In a saturated solution, the rate of dissolution is equal and the same to the rate of crystallization.

  • A saturated solution of as substance (solute) at a particular temperature is one which contains the maximum amount of the substance that can dissolve at that temperature  in the presence of the crystals of the substance.
  • It is an equilibrium system in which a solid substance is in equilibrium with its own ions in solution.
  • Therefore the rate of dissolution will the same with that of crystallization.
8 0
3 years ago
Which of the following properties of a protein is least likely to be affected by changes in pH? Tertiary structure Primary struc
chubhunter [2.5K]

Answer:

Primary structure is the correct answer.

Explanation:

  • The primary structure is the simple level of protein structure.
  • Primary structure is a basic amino acids sequences in a protein.
  • In the primary structure, amino acids are attached together by a covalent bond.
  • Primary structure is when the amino acids are joined together with peptide bonds to produce polypeptide chains
  • Changes in pH are least likely to change the amino acid sequence or disrupt peptide bonds.

Explanation:

7 0
3 years ago
What mass of butane in grams is necessary to produce 1.5×103 kj of heat what mass of co2 is produced?
kari74 [83]
The heat of reaction (i.e. combustion) of butane (C_{4} H_{10}) when reacted with oxygen (O_{2})  is -2658 kJ/mol butane, and the chemical reaction is given by: 

C_{4} H_{10} + \frac{13}{2} O_{2} ---> 4 CO_{2}  + 5 H_{2}O

The mass of butane required in the reaction is based on the heat produced by the reaction, which is given to be -1,500 kJ. The minus sign is added because the reaction releases heat (exothermic), which means that the products are in a "lower energy state" than the reactants. 

Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required. 

Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane

Mass of butane  = 0.5643 moles butane * 58 grams/mol butane
Mass of butane  = 32.73 grams butane

The mass of carbon dioxide (CO_{2}) can be determined by multiplying the moles of butane (C_{4} H_{10}) with the mole ratio of (CO_{2}) produced to the (C_{4} H_{10}) reacted, and then with the molar mass of (CO_{2}), which is 44 grams/mole. 

Mass of carbon dioxide produced 
    = 0.5643 moles butane * [4 moles CO_{2}/ 1 mole C_{4} H_{10}] * 44 grams/mole CO_{2}

Mass of carbon dioxide produced  
    = 99.32 grams CO_{2}

Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams. 
                
4 0
3 years ago
Read 2 more answers
A sample of oxygen gas occupies 3.60 liters at a pressure of 1.00 atm. If temperature is held constant, what will be the volume
Soloha48 [4]

Answer: The volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L

At constant temperature, the volume of a fixed mass of gas is inversely proportional to the pressure it exerts, then

PV = c

Thus, if the pressure increases, the volume decreases, and if the pressure decreases, the volume increases.

It is not necessary to know the exact value of the constant c to be able to use this law since for a fixed amount of gas at constant temperature, it is satisfied that,

P₁V₁ = P₂V₂

Where P₁ and P₂ as well as V₁ and V₂ correspond to pressures and volumes for two different states of the gas in question.

In this case the first oxygen gas state corresponds to P₁ = 1.00 atm and V₁ = 3.60 L while the second state would be P₂ = 2.50 atm and V₂ = y. Substituting in the previous equation,

1.00 atm x 3.60 L = 2.50 atm x y

We cleared y to find V₂,

V₂ = y = \frac{1.00 atm x 3.60 L}{2.50 atm} = 1.44 L

Then, <u>the volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L</u>

3 0
3 years ago
Each liter of air has a mass of 1.80 grams. How many liters of air are contained in
natita [175]

Question:

Each liter of air has a mass of 1.80 grams. How many liters of air are contained in 2.5 x103) kg of air?

Answer:

11.48106 L

5 0
3 years ago
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