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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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(g) On the graph in part (d) , carefully draw a curve that shows the results of the second titration, in which the student titra
atroni [7]

Answer:

Here's what I get  

Explanation:

(g) Titration curves

I can't draw two curves on the same graph, but I can draw two separate curves for you.  

The graph in part (d) had an equivalence point at 20 mL.

In the second titration, the NaOH was twice as concentrated, so the volume to equivalence point would be half as much — 10 mL.

The two titration curves are below.

(h) Evidence of reaction

HCl and NaOH are both colourless.

They don't  evolve a gas or form a precipitate when they react.

The student probably noticed that the Erlenmeyer flask warmed up — a sign of a chemical change.

4 0
2 years ago
Calculate the molarity of 1.0 mol of KCl in 750 ml of solution
Vaselesa [24]

Answer:

[KCl] = 1.33 M

Explanation:

Molarity is mol /L

Mol of solute in 1 L of solution

Volume of solution is 750 mL

750 mL / 1000 = 0.750 L

1 mol / 0.750L = M → 1.33

7 0
2 years ago
Read 2 more answers
What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M methylamine, CH3NH2, with 20.00 mL of 0.10 M methylammonium c
olya-2409 [2.1K]

Answer:

pH=10.97

Explanation:

the solution of methyl amine with methylammonium chloride will make a buffer solution.

The pH of buffer solution can be obtained using Henderson Hassalbalch's equation, which is:

pOH=pKb+log\frac{[salt]}{[base]}

pH = 14- pOH

Let us calculate pOHpOH=3.43+ (-0.397)=3.03

pH=14-pOH=14-3.03=10.97

[Salt] = [methylammonium chloride] = 0.10 M (initial)

After adding base

[salt] = \frac{molarityXvolume}{finalvolume}=\frac{0.1X20}{(20+50)}= 0.0286M

[base] = [Methylamine]=0.10

After mixing with salt

[base]= \frac{molarityXvolume}{finalvolume}=\frac{0.1X50}{(20+50)}= 0.0714M

pKb= -log[Kb]= 3.43

Putting values

pOH = 3.43+log(\frac{[0.0286]}{0.0714}

4 0
2 years ago
You are presented with a mystery as part of your practical experiment. You have a solution of Pb(NO3)2 that has a worn label mak
Orlov [11]

Answer:

Minimum volume of H₂SO₄ required for H₂SO₄ to be in excess = 0.0556 mL

Explanation:

Pb(NO₃)₂ + H₂SO₄ -----> PbSO₄ + 2HNO₃

For this reaction, we know that the max concentration of Pb(NO₃) according to the bottle is 0.999M and to ensure the other reactant in the reaction is in excess, we'll do the calculation with a Pb(NO₃) that's a bit higher, that is, 1.0M.

Knowing that Concentration in mol/L = (number of moles)/(volume in L)

Number of moles of Pb(NO₃) added = concentration in mol/L × volume in L = 1 × 0.001 = 0.001 mole

According to the reaction,

1 mole of Pb(NO₃) reacts with 1 mole of H₂SO₄

0.001 mole of Pb(NO₃) will react with 0.001×1/1 mole of H₂SO₄

Therefore number of H₂SO₄ required for the reaction and for the H₂SO₄ to be in excess is 0.001 mole of H₂SO₄

So, the concentration of commercial H₂SO₄ is usually 18.0M, using this as the assumed value.

Volume of H₂SO₄ = (number of H₂SO₄ required for it to be in excess)/(concentration of H₂SO₄)

Volume of H₂SO₄ = 0.001/18 = 0.0000556 L = 0.0556 mL.

QED!!!

5 0
2 years ago
0.45 g of hydrated sodium carbonate crystals were heated until 3.87 of anhydrous power remained.
snow_tiger [21]

Formula of hydrated sodium carbonate : Na₂CO₃.10H₂O, so moles of water in one mole of hydrated salt = 10

<h3>Further explanation</h3>

Hydrate is a compound that binds water (H₂O), usually in the form of crystals/ solids

If these compounds are dissolved in water or heated, the hydrates can decompose:

Example: X.YH₂O (s) → X (aq) + YH₂O (l)

The formula for the hydrated compound contains: YH2O

The mole ratio shows the ratio of the coefficients of the hydrate compound

10.45 hydrated sodium carbonate(Na₂CO₃.xH₂O) were heated until 3.87 of 3.87of anhydrous (Na₂CO₃) remained, so

mass H₂O released :

\tt 10.45-3.87=6.58~g

mass Na₂CO₃ = 3.87 g

mol ratio Na₂CO₃(MW= g/mol) : H₂O(MW=18 g/mol) =

\tt \dfrac{3.87}{105,9888}\div \dfrac{6.58}{18}=0.0365\div 0.3655=1\div 10

6 0
2 years ago
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