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Pie
3 years ago
10

Which type of bond makes it possible for a macromolecule to interact with great specificity with just one out of the many thousa

nds of different molecules present inside a cell?
Chemistry
1 answer:
Andrei [34K]3 years ago
7 0

Answer:Non-covalent bonds

Explanation:

The Non-covalent bonds are bonds such as van der Waals forces of attraction, the Hydrogen bonds, hydrophobic bonds and so on. The Non-covalent bonds are very important types of bonding in large biological molecules.

Just like the question says, the Non-covalent bonds, ''makes it possible for a macromolecule to interact with great specificity with just one out of the many thousands of different molecules present inside a cell".

Ionic bonding is also a Non-covalent bonding. They(Non-covalent bonds) helps in the stability of large macromolecules.

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Which can be classified as a trace fossil
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It has been hypothesized that a chemical known as BW prevents colds. To test this hypothesis, 20,000 volunteers were divided int
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3 years ago
An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 an
Westkost [7]

Answer:

7.3

Explanation:

By Henderson Hasselbalch equation we can calculate the pH or the pOH of a solution by its pKa. Remember that pH = -log[H^{+}], and pKa = -logKa. Ka is the equilibrium constant of the acid.

Henderson Hasselbalch equation :

pH = pKa - log \frac{[HA]}{[A^{-}]}

Where [HA] is the concentration of the acid, and [A^{-}] is the concentration of the anion which forms the acid.

So, acid X, has two ionic forms, the carboxyl group and the other one. First, we have 0.1 mol/L of the acid, in 100 mL, so the number of moles of X

n1 = (0.1 mol/L)x(0.1 L) = 0.01 mol

When it dissociates, it forms 0.005 mol of the carboxyl group and 0.005 mol of the other group. Assuming same  stoichiometry.

Adding NaOH, with 0.1 mol/L and 75 mL, the number of moles of OH^- will be

n2 = (0.1 mol/L)x(0.075 L) = 0.0075 mol

So, the 0.0075 mol of OH^- reacts with 0.005 mol of carboxyl, remaining 0.0025 mol of OH^-, which will react with the 0.005 mol of the other group. So, it will remain 0.0025 mol of the other group.

The final volume of the solution will be 175 mL, but both concentrations (the acid form and ionic form) have the same volume, so we can use the number of mol in the equation.

Note that, the number of moles of the acid form is still 0.01 mol because it doesn't react!

So,

6.72 = pKa - log \frac{0.01}{0.0025}

6.72 = pKa - log 4

pKa - log4 = 6.72

pKa = 6.72 + log4

pKa = 6.72 + 0.6

pKa = 7.3

8 0
3 years ago
If the pressure on a gas sample is tripled and the absolute temperature is quadrupled, by what factor will the volume of the sam
antoniya [11.8K]

Answer:

b. 4/3

Explanation:

Given data

  • Initial pressure: P₁
  • Initial volume: V₁
  • Initial temperature: T₁
  • Final pressure: P₂ = 3 P₁
  • Final volume: V₂
  • Final temperature: T₂ = 4 T₁

We can find by what factor will the volume of the sample change using the combined gas law.

\frac{P_{1}.V_{1}}{T_{1}} =\frac{P_{2}.V_{2}}{T_{2}}\\\frac{P_{1}.V_{1}}{T_{1}} =\frac{3P_{1}.V_{2}}{4T_{1}}\\V_{1} =\frac{3V_{2}}{4}\\V_{2}=4/3 V_{1}

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3 years ago
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