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Alla [95]
1 year ago
14

The km of an enzyme is 5. 0 mm. Calculate the substrate concentration when this enzyme operates at one‑quarter of its maximum ra

te.
Chemistry
1 answer:
Makovka662 [10]1 year ago
3 0

The substrate concentration of the enzyme operating at one‑quarter of its maximum rate is = 0.333.

Relationship between Km and substrate concentration is -

Km is the concentration of substrate.It allows the enzyme to achieve half Vmax. High Km enzyme requires a higher concentration of substrate to get Vmax. Since, Km is a constant. If the substrate concentration is increased, it has no effect on it.

An enzyme with a high Km has a low affinity for its substrate. The substrate concentration Km corresponds to the substrate concentration.

The substrate concentration at which the reaction rate of the enzyme-catalyzed reaction is half of the maximum reaction rate Vmax.

The equation is:

                <em>V₀ = </em><u><em>Vmax [S] </em></u>

<em>                          [S] + Km</em>

Here,

V₀ is initial rate,

Km is the dissociation constant between the substrate and the enzyme,

Vmax is the maximum rate, and

S is the concentration of substrate.

taking fraction of V₀ and Vmax :

<u><em>    V₀    </em></u><em> = </em><u><em>     [S</em></u><em>]</em><u><em>     </em></u><em>  </em>

<em>Vmax      [S] + Km</em>

<u><em>   </em></u><u>V₀   </u> =   <u>   0.5Km   </u>  = 0.333

Vmax       1.50 + Km

Therefore, the substrate concentration of this enzyme operating at one‑quarter of its maximum rate is = 0.333.

To learn more about substrate concentration,

brainly.com/question/18237939

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<h3>50 kg.m/s</h3>

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3 years ago
N2(g) + 3H2(g) -&gt; 2NH 3(g)
Nutka1998 [239]

Answer:

None of the options are correct. The correct answer is:

56.67g

Explanation:

N2 + 3H2 —> 2NH3

Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

Mass of NH3 from the balanced equation = 2 x 17 = 34g

Molar Mass of H2 = 2x1 = 2g/mol

Mass of H2 from the balanced equation = 3 x 2 = 6g

From the equation,

6g of H2 produced 34g of NH3.

Therefore, 10g of H2 will produce = (10 x 34)/6 = 56.67g of NH3

Therefore, 56.67g of NH3 are produced

5 0
3 years ago
Natural gas is a mixture of many substances, primarily CH₄, C₂H₆, C3H8, and C4₄H₁₀. assuming that the total pressure of the gase
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Answer:

A. The partial pressure for CH4 = 0.0925atm

B. The partial pressure for C2H6 = 0.925atm

C. The partial pressure for C3H8 = 0.346atm

D. The partial pressure for C4H10 = 0.115atm

Explanation:

Total pressure = 1.48atm

Total mole = 0.4+4+1.5+0.5=6.4

A. Mole fraction of CH4 = 0.4/6.4 = 0.0625

The partial pressure for CH4 = 0.0625 x 1.48 = 0.0925atm

B. Mole fraction of C2H6 = 4/6.4 = 0.625

The partial pressure for C2H6 = 0.625 x 1.48 = 0.925atm

C. Mole fraction of C3H8 = 1.5/6.4 = 0.234

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D. Mole fraction of C4H10 = 0.5/6.4 = 0.078

The partial pressure for C4H10 = 0.078 x 1.48 = 0.115atm

7 0
3 years ago
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