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LUCKY_DIMON [66]
3 years ago
11

Why ph changes disrupt protein's ionic bonds?

Chemistry
1 answer:
Inessa05 [86]3 years ago
8 0

Answer:

A change in pH in the protein habitat can modify its ionic bonds because because the chemical equilibrium shifts to one side or the other depends on the modification

Explanation:

The pH influences the charge acquired by the acidic and basic groups present in the molecules. Proteins usually have groups with characteristics of acid or weak base. Therefore, they are partially ionized in solution coexisting in equilibrium different species.

The degree of ionization of the different functional groups is in relation to the pH of the medium in which they are found, since the H3O + and OH- species are part of the equilibrium situation. Therefore, according to the pH, each group with characteristics of weak acid or base present in the molecule will be ionized to a lesser or greater extent. There are extreme situations where the balance has been totally displaced in one direction, for example: under very high pH conditions (low concentration of H3O +) weak acids are considered fully ionized, so the functional group will always have an electric charge. The same goes for the bases at very low pH values. In other equilibrium situations, species of the same molecule with different load will coexist in the solution, due to the pH value of the medium in which it is found.

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You have 16.7 grams of hydrogen and 15.4 grams of oxygen in a synthesis rxn. Which is the limiting reagent?
sleet_krkn [62]

Answer:

oxygen is limiting reactant

Explanation:

Given data:

Mass of hydrogen = 16.7 g

Mass of oxygen = 15.4 g

Limiting reactant = ?

Solution:

Chemical equation:

2H₂ + O₂   →   2H₂O

Number of moles of hydrogen:

Number of moles = mass/ molar mass

Number of moles = 16.7 g/ 2 g/mol

Number of moles = 8.35 mol

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 15.4 g/ 32 g/mol

Number of moles = 0.48 mol

Now we will compare the moles of both reactant with product,

                 

                             H₂           :          H₂O

                              2            :            2

                             8.35        :            8.35

                             O₂           :          H₂O

                               1            :            2

                             0.48        :        2×0.48 = 0.96 mol

The number of moles of water produced by oxygen are less so it will limiting reactant.

3 0
3 years ago
A pure gold bar is made up of only particles. <br><br>A.nickel<br>B.copper <br>C.god<br>D.gold​
Minchanka [31]

Answer: gold

Explanation:

4 0
3 years ago
A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?
Nastasia [14]

Answer:

At -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

                                          \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

Hence    T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

            \Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}

            \Rightarrow T_{2}=260K

            \Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}

So at -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

5 0
3 years ago
The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3 * xH2O.
lesantik [10]

Answer:

44.7 kWh

Explanation:

Let's consider the reduction of Al₂O₃ to Al in the Bayer process.

6 e⁻ + 3 H₂O + Al₂O₃ → 2 Al + 6 OH⁻

We can establish the following relations:

  • The molar mass of Al is 26.98 g/mol.
  • 2 moles of Al are produced when 6 moles of e⁻ circulate.
  • 1 mol of e⁻ has a charge of 96468 c (Faraday's constant).
  • 1 V = 1 J/c
  • 1 kWh = 3.6 × 10⁶ J

When the applied electromotive force is 5.00 V, the energy required to produce 3.00 kg (3.00 × 10³ g) of aluminum is:

3.00 \times 10^{3} gAl.\frac{1molAl}{26.98gAl} .\frac{6mole^{-}}{2molAl}.\frac{96468c}{1mole^{-}}.\frac{5.00J}{c}.\frac{1kWh}{3.6 \times 10^{6}J} =44.7kWh

6 0
3 years ago
Which is a characteristic of atoms?
ra1l [238]

Answer:

the second one ..........

6 0
3 years ago
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