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svp [43]
3 years ago
6

Box 1

Chemistry
2 answers:
Alex73 [517]3 years ago
7 0

10g

Explanation:

Box 1, Mass of A = 10g

Box 2, Mass of B = 5g

Box 3, = 1A + 1B

Unknown:

Mass of B that would combine with mass of 20g of A

Solution:

   Mass ratio of A to B:

   \frac{mass of A}{mass of B} = mass ratio

           \frac{10}{5} = mass ratio

        The mass ratio of A to B = 2: 1

Now, number of B that will combine with 20g of A;

       

           \frac{mass of A}{mass of B} = mass ratio

               \frac{20}{mass of B} = \frac{2}{1}

                     Mass of B = 10g

10g of B would combine with 20g of A

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Vinvika [58]3 years ago
6 0

Answer:

its 10 thats al i got

Explanation:

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enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P . The model can be more readily understood when comparing t
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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

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

_____________________________________________|____________

[ES] is much       |                                 | Almost all active

 lower than         |                                 | sites are filled

[E_{free}]                  |                                 |

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                   [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}]

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So  this means that equation one would now become

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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

             

       

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