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Pachacha [2.7K]
3 years ago
7

Consider all the different polysaccharides that exist across species and then select all of the following characteristics that c

an vary among polysaccharides. a. amount of branching b. types of bonds (glycosidic) between monomers c. types of monosaccharides
Chemistry
1 answer:
sergeinik [125]3 years ago
8 0

Answer:

The correct answer is b types of glycosidic bond between the monomers.

Explanation:

HomoPolysaccharides differ from each other by the presence of different glycosidic linkage within their chemical structure.

For example

1 Starch contain glucose residues which are linked by alpha-1,4-glycosidic linkage.

2 In glycogen molecule the glucose residues are linked together by both alpha-1,4-glycosidic linkage and alpha-1,6- glycosidic linkage.

3  In cellulose the glucose monomers are linked together by beta-1,4-glycosidic linkage.

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Calculate the empirical formula of each of the following substances with the following compositions.
OleMash [197]

Explanation:

Note: Molar masses of elements can be found online or in the periodic table.

Moles of Magnesium

= 3.60g / (24.3g/mol) = 0.148mol.

Moles of Chlorine

= 10.65g / (35.45g/mol) = 0.300mol.

Mole ratio of Magnesium to Chlorine

= 0.148mol : 0.300mol = 1 : 2.

Hence we have the empirical formula MgCl2.

Moles of Lithium

= 9.1g / (6.94g/mol) = 1.311mol.

Moles of Oxygen

= 10.4g / (16g/mol) = 0.650mol.

Moles ratio of Lithium to Oxygen

= 1.311mol : 0.650mol = 2 : 1.

Hence we have the empirical formula Li2O.

7 0
3 years ago
What is the pressure in atm exerted by 2.48 moles of a gas in a 250.0 mL container at 58 degrees celsius
Dvinal [7]
Since the question manages to include moles, pressure, volume, and temperature, then it is evident that in order to find the answer we will have to use the Ideal Gas Equation:  PV = nRT (where P = pressure; V = volume; n = number of moles; R = the Universal Constant [0.082 L·atm/mol·K]; and temperature.

First, in order to work out the questions, there is a need to convert the volume to Litres and the temperature to Kelvin based on the equation:
         250 mL = 0.250 L
             58 °C = 331 K

Also, based on the equation   P = nRT ÷ V

⇒         P  = (2.48 mol)(0.082 L · atm/mol · K)(331 K)  ÷  0.250 L
⇒         P  =  (67.31  L · atm) ÷ 0.250 L
⇒         P  =  269.25 atm

Thus the pressure exerted by the gas in the container is  269.25 atm.
  


7 0
3 years ago
Choose all the answers that apply.
AnnZ [28]

Answer:

accepting your faults

seeing exercise as a treat

looking at your ultimate goal

Explanation:

8 0
3 years ago
A gas under an initial pressure of 0.60 atm is compressed at constant temperature from 27 L to 3.0 L. The final pressure becomes
IRINA_888 [86]

Answer: 5.4

Explanation:

P2 = P1V1/V2

P2 = (.60atm x 27L) / 3.0L  = 5.4atm

8 0
3 years ago
How are solutions related to mixtures?
Ilia_Sergeevich [38]
In chemistry, a solution is a homogeneous mixture composed of two or more substances. In such a mixture, a solute is a substance dissolved in another substance, known as a solvent.
3 0
3 years ago
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