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Alja [10]
2 years ago
12

g In the absence of allosteric effectors, the enzyme phosphofructokinase displays Michaelis–Menten kinetics (see Fig. 7.15). The

v0/Vmax ratio is 0.90 when the concentration of the substrate, fructose-6-phosphate, is 0.10 mM. Calculate the KM for phosphofructokinase under these conditions (in units of mM).
Chemistry
1 answer:
Furkat [3]2 years ago
6 0

Answer:

The value of the Michaelis–Menten constant is 0.0111 mM.

Explanation:

Michaelis–Menten 's equation:

v_o=V_{max}\times \frac{[S]}{(K_m+[S])}=k_{cat}[E_o]\times \frac{[S]}{(K_m+[S])}

V_{max}=k_{cat}[E_o]

Where:

v_o = rate of formation of products

[S] = Concatenation of substrate

[K_m] =  Michaelis constant

V_{max}  = Maximum rate achieved

k_{cat} = Catalytic rate of the system

[E_o] = Initial concentration of enzyme

On substituting all the given values

We have :

\frac{v_o}{V_{max}}=0.90

[S] = 0.10 mM

\frac{v_o}{V_{max}}=\frac{[S]}{(K_m+[S])}

0.90=\frac{0.10 mM}{K_m+0.10 M}

K_m=0.0111 mM

The value of the Michaelis–Menten constant is 0.0111 mM.

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93.2 mL of a 2.03 M potassium fluoride (KF) solution
Marrrta [24]

Answer:

1.98 M

Explanation:

Given data

  • Initial volume (V₁): 93.2 mL
  • Initial concentration (C₁): 2.03 M
  • Volume of water added: 3.92 L

Step 1: Convert V₁ to liters

We will use the relationship 1 L = 1000 mL.

93.2mL \times \frac{1L}{1000mL} = 0.0932 L

Step 2: Calculate the final volume (V₂)

The final volume is the sum of the initial volume and the volume of water.

V_2 = 0.0932L + 3.92 L = 4.01L

Step 3: Calculate the final concentration (C₂)

We will use the dilution rule.

C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{2.03 M \times 3.92L}{4.01L} = 1.98 M

3 0
2 years ago
Read 2 more answers
Titanium metal requires a photon with a minimum energy of 6.94×10⁻¹⁹J to emit electrons. What is the wavelength of this light? E
Elena-2011 [213]

Answer:2.86x10^-7m

Explanation:E=hc/^

E=6.94x10^-19J

c = 2.9979x10^8m/s

h= 6.626x10^-34Js

^ =( 6.626x10^-34)x( 2.9979x 10^8)/ 6.94x10^-19

= 2.86x10^-7m

5 0
2 years ago
A sample of 22K gold contains the following: 22 g gold, 1.0 g silver, and 1.0 g copper. What is the percent gold in the sample?
iris [78.8K]

Answer:

35 percent

Explanation:

8 0
2 years ago
Calculate the heat of reaction, ΔH°rxn, for overall reaction for the production of methane, CH4.
Lesechka [4]

<u>Answer:</u> The enthalpy of the reaction for the production of CH_4 is coming out to be -74.9 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

For the given chemical reaction:

C(s)+2H_2(g)\rightarrow CH_4(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]

We are given:

\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_f_{(H_2)}=0kJ/mol\\\Delta H^o_f_{CH_4}=-74.9kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ

Hence, the enthalpy of the reaction for the production of CH_4 is coming out to be -74.9 kJ

3 0
2 years ago
3. What's the empirical formula of a molecule containing 18.7% lithium, 16.3% carbon, and 65.0% oxygen?
Hunter-Best [27]

Answer:

Empirical formula is  Li₂CO₃.

Explanation:

Percentage of oxygen= 65.0%

Percentage of lithium = 18.7%

Percentage of carbon= 16.3%

Empirical formula = ?

Solution:

Number of gram atoms of C = 16.3/12 = 1.4

Number of gram atoms of Li = 18.7/6.94 = 2.7

Number of gram atoms of O = 65.0/ 16 = 4.1

Atomic ratio:

Li              :            C          :    O

2.7/1.4      :       1.4/1.4         :   4.1/1.4

     2          :            1           :     3  

Li : C : O = 2 : 1 : 3

Empirical formula is  Li₂CO₃.

8 0
2 years ago
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