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

Makovka662 [10]

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:

V₀ = Vmax [S]

[S] + Km

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 :

V₀ = [S]

Vmax [S] + Km

V₀ = 0.5Km = 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,

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3 0

1 year ago

An 85 kg construction worker has 37 485 J of gravitational potential energy. To the nearest whole number, the worker is___

Dmitry [639]

Answer:

44,1 m

Explanation:

Ep= m * g * h

37485 = 85 * 10 * h

h = 44,1 m

7 0

3 years ago

For each process, predict the sign on the entropy change and write a sentence or two to explain how to make this prediction with

MrRa [10]

Answer: a. A solid melts: $\Delta S=+ve$

b. a vapor is converted to solid: $\Delta S=-ve$

c. a liquid freezes: $\Delta S=-ve$

d. A solid sublimes: $\Delta S=+ve$

e. a vapor condenses to liquid : $\Delta S=-ve$

f. a liquid boils: $\Delta S=+ve$

g. dissolving a tablespoon of salt in water: $\Delta S=+ve$

h. combustion of gasoline: $\Delta S=+ve$

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy change is negative and vice versa.

a. A solid melts: The solid is converting to liquid, thus the randomness is increasing as the molecules are moving away and the intermolecular forces are getting weaker. Thus the entropy change $(\Delta S)$ is positive.

b. a vapor is converted to solid: The gas is converting to solid, thus the randomness is decreasing as the molecules are moving close and the intermolecular forces are getting stronger Thus the entropy change $(\Delta S)$ is negative.

c. a liquid freezes: The liquid is converting to solid, thus the randomness is decreasing as the molecules are moving close and the intermolecular forces are getting stronger. Thus the entropy change $(\Delta S)$ is negative.

d. A solid sublimes: The solid is converting to gas, thus the randomness is increasing as the molecules are moving away and the intermolecular forces are getting weaker. Thus the entropy change $(\Delta S)$ is positive.

e. a vapor condenses to liquid : The gas is converting to liquid, thus the randomness is decreasing as the molecules are moving close and the intermolecular forces are getting stronger. Thus the entropy change $(\Delta S)$ is negative.

f. a liquid boils: The liquid is converting to gas, thus the randomness is increasing as the molecules are moving away and the intermolecular forces are getting weaker. Thus the entropy change $(\Delta S)$ is positive.

g. dissolving a tablespoon of salt in water: The solid is converting to ions , thus the randomness is increasing as the ions can move freely. Thus the entropy change $(\Delta S)$ is positive.

h. combustion of gasoline: The liquid is converting to gas, thus the randomness is increasing as the molecules are moving away and the intermolecular forces are getting weaker. Thus the entropy change $(\Delta S)$ is positive.

6 0

2 years ago

PLEASE HELP

Step2247 [10]

Answer:

See explanation

Explanation:

The equation of the reaction is;

C3H8 + 5O2 ----> 3CO2 + 4H2O

Number of moles of C3H8 = 132.33g/44g/mol = 3 moles

1 mole of C3H8 yields 3 moles of CO2

3 moles of C3H8 yields 3 × 3/1 = 9 moles of CO2

Number of moles of oxygen = 384.00 g/32 g/mol = 12 moles

5 moles of oxygen yields 3 moles of CO2

12 moles of oxygen yields 12 × 3/5 = 7.2 moles of CO2

Hence C3H8 is the limiting reactant.

Mass of CO2 produced = 9 moles of CO2 × 44 g/mol = 396 g of CO2

1 moles of C3H8 yields 4 moles of water

3 moles of C3H8 yields 3 × 4/1 = 12 moles of water

Mass of water = 12 moles of water × 18 g/mol = 216 g of water

b) Actual yield = 269.34 g

Theoretical yield = 396 g

% yield = actual yield/theoretical yield × 100/1

% yield = 269.34 g /396 g × 100

% yield = 68%

4 0

3 years ago

You are determining the density of an unknown solid. You figure out the volume by water displacement. Then you weigh the solid,

WINSTONCH [101]

Answer:

The calculated density will be larger

Explanation:

The calculated density will be larger. Because, the volume is taken accurately, by the water displacement method. But, when we the took the mass, the water was present on the unknown solid. So, the mass of that water was added to the original mass of the solid. Hence, the mass measured was larger than the original mass. We, know from the formula of density that density is directly proportional to the mass of the object.

Density = Mass/Volume

Hence, the larger measured mass means the larger value of density.

8 0

3 years ago