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melomori [17]
3 years ago
9

The formation of an enzyme-substrate complex tends to be thermodynamically favorable due to ________ interactions between the su

bstrate and enzyme.
Chemistry
1 answer:
vodomira [7]3 years ago
7 0

Answer:

The correct option is: non-covalent interactions

Explanation:

Enzymes are the macromolecules that are responsible for catalyzing biochemical reactions in the cell and thus they are known as the<u> biological catalysts.</u> They have<u> high specificity and selectivity.</u>

In an enzyme-catalyzed reaction, the substrate molecules bind to the active sites present on the surface of the enzyme, resulting in the formation of a enzyme-substrate complex, which is stabilized by weak non-covalent interactions such as Van der Waals forces, hydrophobic interactions, electrostatic forces and hydrogen bonding.

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Why was mass lost from the crucible during the reaction?
Alexxandr [17]
If the crucible wasn't covered with a lid the reactants may have produced a gas that was released into the surroundings, or mass may have been lost in the form of water vapour.
6 0
3 years ago
if i add 25 ml of water to 135 ml of a 0.25 M NaOH solution what will the molarity of the diluted solution be​
DENIUS [597]

Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

n = c \cdot V,

where

  • n is the number of moles of the solute in the solution.
  • c is the concentration of the solution. c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1} for the initial solution.
  • V is the volume of the solution. For the initial solution, V = 135\;\textbf{mL} = 0.135\;\textbf{L} for the initial solution.

n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}.

What's the concentration of the diluted solution?

\displaystyle c = \frac{n}{V}.

  • n is the number of solute in the solution. Diluting the solution does not influence the value of n. n = 0.03375\;\text{mol} for the diluted solution.
  • Volume of the diluted solution: 25\;\text{mL} + 135\;\text{mL}  = 160\;\textbf{mL} = 0.160\;\textbf{L}.

Concentration of the diluted solution:

\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}.

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

8 0
3 years ago
Read 2 more answers
Please help :)
melamori03 [73]
The driving thrust of the car produced by the engine is the main forward force. 

The main opposing forces are air resistance (from the wind) and friction (between the tyres and the road)

Since the air resistance + friction = driving force the car moves at a constant speed.
4 0
2 years ago
What is the difference between speed and acceleration?<br> Please help :)
ioda

Answer:

Speed is how fast you are going at the moment but acceleration is how fast you are building up speed to get to a set point

Explanation:

4 0
2 years ago
A 1.0 g sample of hydrogen reacts completely with 19.0 g of fluorine to form a compound of hydrogen and fluorine. a. What is the
Rashid [163]

Explanation:tr

a) Molar mass of HF = 20 g/mol

Atomic mass of hydrogen = 1 g/mol

Atomic mass of fluorine = 19 g/mol

Percentage of an element in a compound:

\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{Molar mass of compound }}\times 100

Percentage of fluorine:

\frac{1\times 19 g/mol}{20g/mol}\times 100=95\%

Percentage of hydrogen:

\frac{1\times 1g/mol}{20 g/mol}\times 100=5\%

b) Mass of hydrogen in 50 grams of HF sample.

Moles of HF = \frac{50 g}{20 g/mol}=2.5 mol

1 mole of HF has 1 mole of hydrogen atom.

Then 2.5 moles of HF will have:

1\times 2.5 mol=2.5 mol of hydrogen atom.

Mass of 2.5 moles of hydrogen atom:

1 g/mol × 2.5 mol = 2.5 g

2.5 grams of hydrogen would be present in a 50 g sample of this compound.

c) As we solved in part (a) that HF molecules has 5% of hydrogen by mass.

Then mass of hydrogen in 50 grams of HF compound we will have :

5% of 50 grams of HF = \frac{5}{100}\times 50 g=2.5 g

8 0
3 years ago
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