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Anna [14]
3 years ago
8

(a) Glutenin, a wheat protein rich in disulfide bonds, is responsible for the cohesive and elastic character of dough made from

wheat flour. Similarly, the hard, tough nature of tortoise shell is due to the extensive disulfide bonding in its -keratin. What is the molecular basis for the correlation between disulfide-bond content and mechanical properties of the protein?
Chemistry
1 answer:
lilavasa [31]3 years ago
4 0

Answer: Through extensive bonding

Explanation:

Oxidation of two cysteine gives disulphide bonds. They are covalent in nature. Given as;

R-CH2-SH + R'-CH2 +02 = R-CH2-S-S-CH2-R' + H202

They are the stabilizing force behind protein folding the endoplasmic reticulum. Proteins being the backbone of our genetic constitution require disulphide bonding to stabilize the peptide molecules and the extent and number of this bonding determines the Resistance of the protein molecule to extreme conditions e.g extreme heat which denatures proteins and elasticity as well.

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The following table lists molecular weight data for a polypropylene material. Molecular Weight Range (g/mol) xi wi
nikklg [1K]

Answer:

a) the number-average molecular weight is 32,400 g/mol

b) the weight-average molecular weight is 36,320 g/mol

c) the degree of polymerization is 770

Explanation:

Given the data in the question;

Molecular                  Mean        Number      Weight      Number        Weight

weight                                                                             average         average

(g/mol)                    Mi(g/mol)         xi                wi          xiMi                  wiMi

8,000-16,000            12,000        0.07           0.03         840                360

16,000-24,000          20,000       0.15           0.09         3000              1800

24,000-32,000         28,000       0.26           0.21          7280              5880

32,000-40,000         36,000       0.27           0.27         9720               9720

40,000-48,000         44,000       0.18            0.28         7920              12320

48,000-56,000         52,000       0.07           0.12          3640              6240

∑                                                                                Mn=32,400     Mw=36,320

so;

a)  the number-average molecular weight

Mn = ∑Mixi

so from the table above; summation of Row Mixi

Mn = ∑Mixi = 32,400

Therefore, the number-average molecular weight is 32,400 g/mol

b) the weight-average molecular weight

Mw = ∑Miwi

so from the table above; summation of Row Miwi

Mw = ∑Miwi = 36,320

Therefore, the weight-average molecular weight is 36,320 g/mol

c) the degree of polymerization

the degree of polymerization of polypropylene can be determined using number-average molecular and repeat unit molecular weight.

now, for polypropylene { CH₂ = CH - CH₃ }

the repeat unit consist of 3 carbon atoms and 6 hydrogen atoms

given that;

Atomic weight of Carbon mC = 12.01 g/mol and

Atomic weight of  Hydrogen mH = 1.008 g/mol

now we find the repeat unit molecular weight of polypropylene

m = nCmC + nHmH

where n is the number of repeat of atoms

so we substitute

m = ( 3 × 12.01) + ( 6 × 1.008)

m = 36.03 + 6.048

m = 42.078 g/mol

now we calculate the degree of polymerization;

DP = Mn / m

so we substitute

DP = 32,400 / 42.078

DP = 769.9985 ≈ 770

Therefore,  the degree of polymerization is 770

3 0
3 years ago
Which description of salt is a physical property?
Nonamiya [84]

Answer:

White being the color and coming in small grains.

Explanation:

Physical properties are something you can clearly see about the object.

6 0
2 years ago
Read 2 more answers
A compound has the molecular formula C3H7OH. Which class of organic compounds does it belong to?
coldgirl [10]

Hi

Please find attached file with answers.

Hope it help!


Download docx
8 0
3 years ago
Read 2 more answers
When 1.045 g of CaO is added to 50.0 mL of water at 25.0 °C in a calorimeter, the temperature of the water increases to 32.3 °C.
Rus_ich [418]

Answer:

1.71 kJ/mol

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

\Delta H=m\times C\times \Delta T

Where,  

\Delta H  is the enthalpy change

m is the mass

C is the specific heat capacity

\Delta T  is the temperature change

Thus, given that:-

Mass of CaO = 1.045 g

Specific heat = 4.18 J/g°C

\Delta T=32.3-25.0\ ^0C=7.3\ ^0C

So,  

\Delta H=1.045\times 4.18\times 7.3\ J=31.88713\ J

Also, 1 J = 0.001 kJ

So,  

\Delta H=0.03189\ kJ

Also, Molar mass of CaO = 56.0774 g/mol

Moles=\frac{Mass}{Molar\ mass}=\frac{1.045}{56.0774}\ mol=0.01863\ mol

Thus, Enthalpy change in kJ/mol is:-

\Delta H=\frac{0.03189}{0.01863}\ kJ/mol=1.71\ kJ/mol

8 0
3 years ago
A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and
Romashka [77]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

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

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (S_8) = 0.800 g

M_{solute} = Molar mass of solute (S-8) = 256.52 g/mol

W_{solvent} = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m

  • <u>Calculation for freezing point of solution:</u>

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

or,

\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC

Hence, the freezing point of solution is 16.5°C

  • <u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

or,

\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC

Hence, the boiling point of solution is 118.2°C

8 0
3 years ago
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