What is fractional distinction

The Michaelis-Menten equation is an expression of the relationship between the initial velocity,V0, of an enzymatic reaction and

antiseptic1488 [7]

Answer:

Option 2, Half of the active sites are occupied by substrate

Explanation:

Michaelis-Menten expression for enzyme catalysed equation is as follows:

$V_0=\frac{V_{max\ [S]}}{k_M+[S]}$

Here, $K_m$ is Michaelis-Menten constant and [S] is substrate concentration.

When [S]=Km

Rearrange the above equation as follows:

$V_0=\frac{V_{max}[S]}{k_M+[S]}\\V_0=\frac{V_{max}[S]}{[S]+[S]}\\V_0=\frac{V_{max}[S]}{2[S\\]}\\V_0=\frac{V_{max}}{2}$

when [S]=Km, the rate of enzyme catalysed reaction becomes half of the maximum rate, that means half of the active sites are occupied by substrate.

Therefore, the correct option is option 2.

4 0

4 years ago

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likoan [24]

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<h3><u>First State</u> : Punjab </h3>

<u>soil</u> - Alluvial soil

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<u>soil</u> - black soil

<u>Crops</u> - Cotton is the major crop. Other crops include wheat, cereals, rice, jowar, sugarcane, linseed,

<u>Suitable properties</u> : Black soils get sticky when fully wet. Rich in magnesium, iron, aluminum, and lime. They moisture excellently.

<h3><u>Third state</u> : Orrisa </h3>

<u>soil</u> - laterite soil

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<u>suitable properties</u> : These Soils are acidic in nature and coarse and crumbly in texture. it is also rich in iron content.

<h3><u>Fourth state</u> : Rajasthan </h3>

<u>soil</u> - Arid soil

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<h3><u>Fifth state</u> : Sikkim </h3>

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8 0

2 years ago

Read 2 more answers

Vinegar, the commercial name for acetic acid, HC2Hs02, is a monoprotic organic acid. A 5% (w/v) solution of vinegar is used to t

damaskus [11]

19.2 g of vinegar solution

Explanation:

Here we have the chemical reaction between acetic acid (CH₃COOH) and calcium carbonate (CaCO₃):

2 CH₃COOH + CaCO₃ → (CH₃COO)₂Ca + CO₂ + H₂O

number of moles = mass / molecular weight

number of moles of CaCO₃ = 0.8 / 100 = 0.008 moles

Knowing the chemical reaction, we devise the following reasoning:

if 2 moles of CH₃COOH react with 1 moles of CaCO₃

then X moles of CH₃COOH react with 0.008 moles of CaCO₃

X = (2 × 0.008) / 1 = 0.016 moles of CH₃COOH

mass = number of moles × molecular weight

mass of acetic acid (CH₃COOH) = 0.016 × 60 = 0.96 g

Now to find the volume of vinegar acid (solution of acetic acid) with a concentration of 5% (weight/volume) we use the following reasoning:

if there are 5 g of acetic acid in 100 mL of vinegar solution

then there are 0.96 g of acetic acid in Y mL of vinegar solution

Y = (0.96 × 100) / 5 = 19.2 g of vinegar solution

Learn more about:

weight/volume concentration

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8 0

3 years ago

Consider the following Reaction.

Evgen [1.6K]

Answer: -

O 2 limiting reagent

53.83 g CO2 theoretical Yield

97.9% percentage yield

Explanation: -

Mas of CH4 = 23.2 g

Molar mass of CH4 = 12 x 1 + 1 x 4 = 16g

Mass of O2 = 78.3 g

Molar mass of O 2 – 16 x 2 = 32 g

The balanced chemical equation for the reaction is

CH4 + 2 O2 = CO2 + 2 H2O

From the balanced equation we see that

2 O 2 reacts with 1 CH4

2 x 32 g of O 2 react with 16 g of CH4

78.3 g of O 2 react with $\frac{16 g CH4 x 78.3 g O2}{2 x32 g O2}$

= 19.575 g pf CH4

Thus CH4 is in excess.

The limiting reagent is thus O 2.

Molar mass of CO2 = 12 x 1 + 16 x 2= 12 +32 = 44g

From the balanced equation we see

2O 2 gives 1 CO2

2 x 32 g of O 2 gives 44g of CO2

78.3 g of O 2 gives = $\frac{44 gCO2 x 78.3 g O2}{2 x 32 g O2}$

=53.83 g of CO2

Theoritical Yield = 53.83 g of CO2

Actual yield = 52.7 g

Percentage yield = $\frac{52.7 g}{53.83 g} x 100$

=97.9 %

5 0

3 years ago

what is the mass of a sample of o2 gas, which has a pressure of 740mmhg, at a temperature of 25c, in a volume of 250.l

lions [1.4K]

From PV=nRT whereby n= mass per molar mass therefore PV=mRT over Mr making subject m it will be m =PV over RT substuting the above data 25+273 in kelvin 298k then m=740x 250/ 0.0821 x 298 multiply and divide the above eqn u will get the answer

5 0

3 years ago

What is fractional distinction​

What is fractional distinction