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kirill115 [55]
3 years ago
10

Consider the reaction. Upper H subscript 2 upper o (g) plus upper C l subscript 2 upper O (g) double-headed arrow 2 upper H uppe

r C l upper O (g). At equilibrium, the concentrations of the different species are as follows. [H2O] = 0.077 M [Cl2O] = 0.077 M [HClO] = 0.023 M What is the equilibrium constant for the reaction at this temperature?
Chemistry
2 answers:
Vlad [161]3 years ago
6 0

Answer:

0.089

Explanation:

Step 1:

The balanced equation for the reaction is given below:

H2O + Cl2O <=> 2HClO

Step 2:

Data obtained from the question. This includes:

Concentration of H2O, [H2O] = 0.077 M

Concentration of Cl2O, [Cl2O] = 0.077 M

Concentration of HClO, [HClO] = 0.023 M

Equilibrium constant, K =?

Step 3:

Determination of the equilibrium constant. This is illustrated below:

The equilibrium constant for the above reaction is given below:

K = [HClO]^2 / [H2O] [Cl2O]

K = (0.023)^2 / (0.077 x 0.077)

K = 0.089

Therefore, the equilibrium constant for the above reaction is 0.089

dedylja [7]3 years ago
5 0

Answer:

Equilibrium constant for the reaction is 0.0892

Explanation:

This is the reaction of equilibrium

         H₂O      +     Cl₂O     ⇄    2HClO

Eq   0.077M       0.077M           0.023M

Let's make the expression for Kc

Kc = [HClO]² / [Cl₂O] . [H₂O]

Kc= 0.023² / 0.077 . 0.077 = 0.0892

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Answer:

The K_{m} of a substrate will be "10 μM".

Explanation:

The given values are:

E_{t} = 20 \ nM

[Substract] = 40 \ \mu M

K_{cat}=600 \ s^{-1}

Reaction velocity, Vo=9.6 \ \mu M s^{-1}

As we know,

⇒  Vo=\frac{K_{cat}[E_{t}][S]}{K_{m}+[S]}

On putting the estimated values, we get

⇒  9.6=\frac{600\times 20\times 10^{-3}\times 40}{K_{m}+40}

⇒  K_{m}+40=\frac{600\times 20\times 10^{-3}\times 40}{9.6}

⇒  K_{m}+40=50

On subtracting "40" from both sides, we get

⇒  K_{m}+40-40=50-40

⇒  K_{m}=10 \ \mu M

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3 years ago
What is the correct order of steps to determine the mass of product created, given a certain mass of reactant?
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Explanation:

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3 years ago
I need help with 1,2,3, and 4
Schach [20]

Answer:

  • Problem 1: 1.85atm
  • Problem 2: 110mL
  • Problem 3: 290 mL
  • Problem 4: 1.14 atm

Explanation:

Problem 1

<u>1. Data</u>

<u />

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

<u>2. Formula</u>

Since the temeperature is constant you can use Boyle's law for idial gases:

          PV=constant\\\\P_1V_1=P_2V_2

<u>3. Solution</u>

Solve, substitute and compute:

         P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2

        P_2=3.25atm\times755mL/1325mL=1.85atm

Problem 2

<u>1. Data</u>

<u />

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

<u>2. Formula</u>

You assume that the temperature does not change, and then can use Boyl'es law again.

          P_1V_1=P_2V_2

<u>3. Solution</u>

This time, solve for V₂:

           P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2

Substitute and compute:

        V_2=548mmHg\times 125mL/625mmHg=109.6mL

You must round to 3 significant figures:

        V_2=110mL

Problem 3

<u>1. Data</u>

<u />

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

<u>2. Formula</u>

At constant pressure, Charle's law states that volume and temperature are inversely related:

         V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

The temperatures must be in absolute scale.

<u />

<u>3. Solution</u>

a) Convert the temperatures to kelvins:

  • T₁ = 25 + 273.15K = 298.15K

  • T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

        \dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml

You must round to two significant figures: 290 ml

Problem 4

<u>1. Data</u>

<u />

a) P = 865mmHg

b) Convert to atm

<u>2. Formula</u>

You must use a conversion factor.

  • 1 atm = 760 mmHg

Divide both sides by 760 mmHg

       \dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}

<u />

<u>3. Solution</u>

Multiply 865 mmHg by the conversion factor:

    865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer

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