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Alekssandra [29.7K]
3 years ago
10

What happens to the average kinetic energy of a gas when the particles of the gas collide against each other at a constant tempe

rature and volume? The average kinetic energy remains constant. The average kinetic energy drastically increases. The average kinetic energy drastically decreases. The average kinetic energy gets added to the potential energy.
Chemistry
2 answers:
Triss [41]3 years ago
8 0

Answer:

the average kinetic energy remains constant

Explanation:

Collisions between gas particles are elastic. An elastic collision means there is no net gain or loss of kinetic energy due to the collision. This means a sample of gas will have a constant average kinetic energy at a constant temperature and volume.

LuckyWell [14K]3 years ago
5 0

Answer:

The average kinetic energy drastically increases.

Explanation:

The average kinetic energy drastically increases when the gas particles collide against each other at a constant temperature and volume because the collision will increase the heat content of gas at a particular temperature and gas particles begin to move faster as gain more kinetic energy.

Hence, the correct answer is "B".

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3 years ago
enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P . The model can be more readily understood when comparing t
laila [671]

Complete Question

The complete question is shown on the first uploaded image

Answer:

[S]<<KM             |   [S]=KM                  |  [S]>>KM                     | Not true

____________  |   Half of the active  | Reaction rate is         | Increasing

[E_{free}] is about   |    sites are filled of  |    independent of      |  [E_{Total}] will                                            

 equal to [E_{total}]. |                                 |   [S]                             | lower KM

_____________________________________________|____________

[ES] is much       |                                 | Almost all active

 lower than         |                                 | sites are filled

[E_{free}]                  |                                 |

Explanation:

Generally the combined enzyme[ES] is mathematically represented as

                   [ES] = \frac{[E_{total}][S]}{K_M + [S]}----(1)

for Michaelis-Menten equation

Where [S] is the substrate concentration and K_M is the Michaelis constant

Considering the statement [S] < < K_M

  Looking at the equation [S] is denominator so it can be ignored(it is far too small compared to K_M)  hence the above equation becomes

               [ES] = \frac{[E_{total}][S]}{K_M}

Since [S] is less than K_M it means that \frac{[S]}{K_M}  < < 1

so it means that [ES] < < [E_{total}]

  What this means is that the  number of combined enzymes[ES] i.e the number of occupied site is very small compared to the the total sites [E_{total}]  i.e the total enzymes concentration which means that the free sites [E_{free}]  i.e the concentration of free enzymes is almost equal to [E_{total}]

Considering the second statement

      [S] = K_M

So  this means that equation one would now become

           [ES] = \frac{[E_{total}][S]}{2[S]} = \frac{[E_{total}]}{2}

So this means that half of the active sites that is the total enzyme concentration are filled with S

Considering the Third Statement

      [S] >>K_M

In this case the K_M in the denominator of equation 1 would be neglected and the equation becomes

       [ES] = \frac{[E_{total}] [S]}{[S]} = [E_{total}]

This means that almost all the sites are occupied with substrate

 The rate of this reaction is mathematically defined as

             v =\frac{V_{max}[S]}{K_M [S]}

Where v is the rate of the reaction(also know as the velocity of the reaction at a given time t) and V_{max}  is he maximum velocity of the reaction

In this case also the K_M at the denominator would be neglected as a result of the statement hence the equation becomes

                v = \frac{V_{max}[S]}{[S]} = V_{max}

So it means that the reaction does not depend on the concentration of substrate [S]

For the final statement(Not True ) it would match with condition that states that increasing [E_{total}] will lower K_M

This is because K_M does not depend on enzyme concentration it is a property of a enzyme

             

       

7 0
3 years ago
How many ml of 2.50 M NaOH solution are required to make a 525 mL of 0.150 M?
larisa [96]

Answer:

31.5 mL of a 2.50M NaOH solution

Explanation:

Molarity (M) is an unit of concentration defined as moles of solute (In this case, NaOH), per liter of solvent. That is:

Molarity = moles solute / Liter solvent

If you want to make 525mL (0.525L) of a 0.150M of NaOH, you need:

0.525L × (0.150mol / L) = <em>0.07875 moles of NaOH</em>

<em />

If you want to obtain these moles from a 2.50M NaOH solution:

0.07875mol NaOH × (1L / 2.50M) = 0.0315L = <em>31.5 mL of a 2.50M NaOH solution</em>

7 0
3 years ago
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