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Deffense [45]
3 years ago
14

A biologist wants to increase the rate of his chemical reaction but has a limited amount of enzyme. He continues to increases th

e substrate concentration instead. Eventually, the reaction rate levels off, and he can't get it to go any faster. What prevented the rate from increasing further?
The solution ran out of enzyme
The substrate concentration reached Vmax
The products of the reaction inhibited the enzyme
The solution ran out of reactants
Physics
1 answer:
boyakko [2]3 years ago
3 0

Answer:

The substrate concentration reached Vmax

Explanation:

In this scenerio, further increase in the rate of reaction was prevented because the substrate concentration hit Vmax.

Beyond the Vmax, the substrate can no longer proceed.

  • The substrate is the reactant on which the enzyme works on.
  • Enzymes are natural organic chemical  catalysts.
  • Increasing the concentration of reactants is one of the known and proven ways to speed up the rate of reaction.
  • Since we have limited amount of enzymes, once they all bound to the substrate, no further increase in the substrate will have an effect on the rate.
  • At the optimum reaction point, all the available substrate will bond with the enzyme.
  • Beyond this, the enzymes are saturated will not further any reaction.
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3 years ago
What is the surface temperature of Sirius B in Kelvins?
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2 years ago
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¿Por qué si cargas a uno de tus compañeros por cierto tiempo no estás realizando un trabajo mecánico?
VladimirAG [237]

Answer:

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Why if you charge a mate by an amount of time you are not doing work?

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4 0
4 years ago
If 500g of water at 20 degrees C is mixed with 750g of water at 30 degrees C, what will the temperature of the mixture be?
hjlf

Answer:

26 ^\circ C

Explanation:

Given that the temperature of  500g of water and 750 g of water are

at 20^{\circ}C and 30^\circ C respectively.

Let m_1=500g, T_1= 20^\circ C

and m_2=750g, T_2= 30^\circC.

The specific heat capacity of water is,

C= 4.186 J/g ^\circ C.

Let the final temperature of the mixture be T^\circ C.

As there is no energy loss, so, the energy loss by the water at higher temperature, i.e. 30^\circ C, will be equal to the energy gain by the water at lower temperature, i.e. 20^{\circ}C.

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\Rightarrow m_1T+m_2T=m_1T_1+m_2T_2

\Rightarrow T=\frac{m_1T_1+m_2T_2}{m_1+m_2}

Now, putting the given value in the above equation, we have

\Rightarrow T=\frac {500\times 20+750\times 30}{500+750}

\Rightarrow T=26^\circ C.

Hence, the temperature of the mixture will be 26 ^\circ C.

5 0
3 years ago
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Explanation:

I copied and pasted this so you dont have to use this answer but im just being truthful.... unlike some people!

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