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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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The equilibrium constant, Kc, for the reaction H2 (g) + I2 (g) ⇄ 2HI (g) at 425°C is 54.8. A reaction vessel contains 0.0890 M H
Mars2501 [29]

Answer: The reaction is not at equilibrium and will proceed to make more products to reach equilibrium.

Explanation:  

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{eq}

K is the constant of a certain reaction when it is in equilibrium, while Q is the reaction  quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For the given chemical reaction:

H_2(g)+I_2(g)\rightarrow 2HI(g)

The expression for Q is written as:

Q=\frac{[HI]^2}{[H_2]^1[I_2]^1}

Q=\frac{[0.0890]^2}{[0.215]^1[0.498]^1}

Q=0.074

Given : K_{eq} = 54.8

Thus as Q, the reaction will shift towards the right i.e. towards the product side.

4 0
2 years ago
Henri becquerel and the curies found out that _____.
natulia [17]

Henri becquerel and the curies found out that atoms were not indivisible and indestructible

Explanation:

The French Physicist, Henri Becquerel, created the appearance of radioactivity. He shared the Nobel Prize in Physics with Marie Curie and Pierre Curie (Marie's husband) in 1903 for their work in radiation. A radionuclide will emit transmission through the process of radioactive sense.

5 0
2 years ago
Read 2 more answers
If you have 10.0 grams of citric acid with enough baking soda (nahco3 how many moles of carbon dioxide can you produce?
kap26 [50]
Easy stoichiometry conversion :)

So, for stoichiometry, we always start with our "given". In this case, it would be the 10.0 grams of NaHCO3. This unit always goes over 1.

So, our first step would look like this:

10.0
------
  1

Next, we need to cancel out grams to get to moles. To do this, we will do grams of citric acid on the BOTTOM of the next step, so it cancels out. This unit in grams will be the mass of NaHCO3, which is 84.007. Then, we will do our unit of moles on top. Since this is unknown, it will be 1.

So, our 2nd step would look like this:

1 mole CO2
-----------------
84.007g NaHCO3

When we put it together: our complete stoichiometry problem would look like this:

10.0g NaHCO3     1mol CO2
---------------------- x -------------------------
            1                  84.007g NaHCO3

Now to find our answer, all we need to do is:
Multiply the two top numbers together (which is 10.0)
Multiply the two bottom numbers together (Which is 84.007)

And then....

Divide the top answer by the bottom answer.

10.0/84.007 is 0.119

So, from 10.0 grams of citric acid, we have 0.119 moles of CO2.

Hope I could help!
6 0
2 years ago
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Which of the following solution is more dilute and explain why?a)1M b)2M c)0.1M or d)0.009M
Taya2010 [7]
I can’t access the pictures. Sorry!
6 0
2 years ago
How many moles of gas Does it take to occupy 520 mL at a pressure of 400 torr and a temperature of 340 k
Ann [662]
Answer would be B. I provided work on an image attached. Message me if u have any other questions on how to do it

6 0
2 years ago
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