A Plot of μ v s . S for S between 0.0 a n d 1.0 k g / m 3 has to be drawn. Concept Introduction: The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation: μ = μ max S K s + S Where, μ = first-order growth rate (s -1 ) μ max = M a x i m u m g r o w t h r a t e S = S u b s t r a t e c o n c e n t r a t i o n K s = the value of S that gives one-half of the maximum growth rate (in kg/m 3 )

Question

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Answer 1

Question

Chapter 16, Problem 16.121P

(a)

Interpretation Introduction

Interpretation:

A Plot of μ vs. S for S between 0.0 and 1.0 kg/m3 has to be drawn.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

(a)

Expert Solution

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.00 to 1.0 kg/m3Ks= 0.03 kg/m3

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.25 kg/m3)(0.03kg/m3)+ (0.25 kg/m3) = 1.34×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(0.50 kg/m3)(0.03kg/m3)+ (0.50 kg/m3) = 1.42×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(0.75 kg/m3)(0.03kg/m3)+ (0.75 kg/m3) = 1.44×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(1.0 kg/m3)(0.03kg/m3)+ (1.0 kg/m3) = 1.46×10−4s−1

S(kg/m3)μ(10−4s−1)0.251.340.501.420.751.441.001.46

By taking all the above data, a graph between μ and S can be plotted like below,

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 16, Problem 16.121P

Figure.1

(b)

Interpretation Introduction

Interpretation:

The density after one hour if the initial S is 0.30 kg/m3 with given initial population density of 5.0×103 cells/m3 has to be determined.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

The first-order reaction kinetics,

ln[A]t= ln[A]0 + k0t

Where,

[A]t= Concentration of reactant at time t[A]0= Initial Concentration of reactantk0= First-order rate constant

(b)

Expert Solution

Answer to Problem 16.121P

The density after one hour has been determined as 8.2×103 cells/m3.

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.30 kg/m3Ks= 0.03 kg/m3t= 1h=3600 s

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.25 kg/m3)(0.03kg/m3)+ (0.25 kg/m3) = 1.363×10−4s−1.

According to the first-order kinetics,

ln[A]t= ln[A]0 + k0tln[A]t= ln[5.0×103] + (1.363×10−4s−1)(3600 s)ln[A]t= 9.00808919[A]t= 8.1689×103 = 8.2×103 cells/m3.

Therefore, the density after one hour is found out to be 8.2×103 cells/m3.

(c)

Interpretation Introduction

Interpretation:

The density after one hour if the initial S is 0.70 kg/m3 has to be determined.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

The first-order reaction kinetics,

ln[A]t= ln[A]0 + k0t

Where,

[A]t= Concentration of reactant at time t[A]0= Initial Concentration of reactantk0= First−order rate constant

(c)

Expert Solution

Answer to Problem 16.121P

The density after one hour if the initial S is 0.70 kg/m3 has been determined as 8.4×103 cells/m3.

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.70 kg/m3Ks= 0.03 kg/m3t= 1h=3600 s

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.70 kg/m3)(0.03kg/m3)+ (0.70 kg/m3) = 1.438356×10−4s−1.

According to the first-order kinetics,

ln[A]t= ln[A]0 + k0tln[A]t= ln[5.0×103] + (1.438356×10−4s−1)(3600 s)ln[A]t= 9.03500135[A]t= 8.39172×103 = 8.4×103 cells/m3.

Therefore, the density after one hour is found out to be 8.4×103 cells/m3.

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Answer 2

Question

Chapter 16, Problem 16.121P

(a)

Interpretation Introduction

Interpretation:

A Plot of μ vs. S for S between 0.0 and 1.0 kg/m3 has to be drawn.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

(a)

Expert Solution

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.00 to 1.0 kg/m3Ks= 0.03 kg/m3

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.25 kg/m3)(0.03kg/m3)+ (0.25 kg/m3) = 1.34×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(0.50 kg/m3)(0.03kg/m3)+ (0.50 kg/m3) = 1.42×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(0.75 kg/m3)(0.03kg/m3)+ (0.75 kg/m3) = 1.44×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(1.0 kg/m3)(0.03kg/m3)+ (1.0 kg/m3) = 1.46×10−4s−1

S(kg/m3)μ(10−4s−1)0.251.340.501.420.751.441.001.46

By taking all the above data, a graph between μ and S can be plotted like below,

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 16, Problem 16.121P

Figure.1

(b)

Interpretation Introduction

Interpretation:

The density after one hour if the initial S is 0.30 kg/m3 with given initial population density of 5.0×103 cells/m3 has to be determined.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

The first-order reaction kinetics,

ln[A]t= ln[A]0 + k0t

Where,

[A]t= Concentration of reactant at time t[A]0= Initial Concentration of reactantk0= First-order rate constant

(b)

Expert Solution

Answer to Problem 16.121P

The density after one hour has been determined as 8.2×103 cells/m3.

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.30 kg/m3Ks= 0.03 kg/m3t= 1h=3600 s

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.25 kg/m3)(0.03kg/m3)+ (0.25 kg/m3) = 1.363×10−4s−1.

According to the first-order kinetics,

ln[A]t= ln[A]0 + k0tln[A]t= ln[5.0×103] + (1.363×10−4s−1)(3600 s)ln[A]t= 9.00808919[A]t= 8.1689×103 = 8.2×103 cells/m3.

Therefore, the density after one hour is found out to be 8.2×103 cells/m3.

(c)

Interpretation Introduction

Interpretation:

The density after one hour if the initial S is 0.70 kg/m3 has to be determined.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

The first-order reaction kinetics,

ln[A]t= ln[A]0 + k0t

Where,

[A]t= Concentration of reactant at time t[A]0= Initial Concentration of reactantk0= First−order rate constant

(c)

Expert Solution

Answer to Problem 16.121P

The density after one hour if the initial S is 0.70 kg/m3 has been determined as 8.4×103 cells/m3.

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.70 kg/m3Ks= 0.03 kg/m3t= 1h=3600 s

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.70 kg/m3)(0.03kg/m3)+ (0.70 kg/m3) = 1.438356×10−4s−1.

According to the first-order kinetics,

ln[A]t= ln[A]0 + k0tln[A]t= ln[5.0×103] + (1.438356×10−4s−1)(3600 s)ln[A]t= 9.03500135[A]t= 8.39172×103 = 8.4×103 cells/m3.

Therefore, the density after one hour is found out to be 8.4×103 cells/m3.

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Answer 3

Question

Chapter 16, Problem 16.121P

(a)

Interpretation Introduction

Interpretation:

A Plot of μ vs. S for S between 0.0 and 1.0 kg/m3 has to be drawn.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

(a)

Expert Solution

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.00 to 1.0 kg/m3Ks= 0.03 kg/m3

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.25 kg/m3)(0.03kg/m3)+ (0.25 kg/m3) = 1.34×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(0.50 kg/m3)(0.03kg/m3)+ (0.50 kg/m3) = 1.42×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(0.75 kg/m3)(0.03kg/m3)+ (0.75 kg/m3) = 1.44×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(1.0 kg/m3)(0.03kg/m3)+ (1.0 kg/m3) = 1.46×10−4s−1

S(kg/m3)μ(10−4s−1)0.251.340.501.420.751.441.001.46

By taking all the above data, a graph between μ and S can be plotted like below,

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 16, Problem 16.121P

Figure.1

(b)

Interpretation Introduction

Interpretation:

The density after one hour if the initial S is 0.30 kg/m3 with given initial population density of 5.0×103 cells/m3 has to be determined.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

The first-order reaction kinetics,

ln[A]t= ln[A]0 + k0t

Where,

[A]t= Concentration of reactant at time t[A]0= Initial Concentration of reactantk0= First-order rate constant

(b)

Expert Solution

Answer to Problem 16.121P

The density after one hour has been determined as 8.2×103 cells/m3.

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.30 kg/m3Ks= 0.03 kg/m3t= 1h=3600 s

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.25 kg/m3)(0.03kg/m3)+ (0.25 kg/m3) = 1.363×10−4s−1.

According to the first-order kinetics,

ln[A]t= ln[A]0 + k0tln[A]t= ln[5.0×103] + (1.363×10−4s−1)(3600 s)ln[A]t= 9.00808919[A]t= 8.1689×103 = 8.2×103 cells/m3.

Therefore, the density after one hour is found out to be 8.2×103 cells/m3.

(c)

Interpretation Introduction

Interpretation:

The density after one hour if the initial S is 0.70 kg/m3 has to be determined.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

The first-order reaction kinetics,

ln[A]t= ln[A]0 + k0t

Where,

[A]t= Concentration of reactant at time t[A]0= Initial Concentration of reactantk0= First−order rate constant

(c)

Expert Solution

Answer to Problem 16.121P

The density after one hour if the initial S is 0.70 kg/m3 has been determined as 8.4×103 cells/m3.

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.70 kg/m3Ks= 0.03 kg/m3t= 1h=3600 s

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.70 kg/m3)(0.03kg/m3)+ (0.70 kg/m3) = 1.438356×10−4s−1.

According to the first-order kinetics,

ln[A]t= ln[A]0 + k0tln[A]t= ln[5.0×103] + (1.438356×10−4s−1)(3600 s)ln[A]t= 9.03500135[A]t= 8.39172×103 = 8.4×103 cells/m3.

Therefore, the density after one hour is found out to be 8.4×103 cells/m3.

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Answer 4

Question

Chapter 16, Problem 16.121P

(a)

Interpretation Introduction

Interpretation:

A Plot of μ vs. S for S between 0.0 and 1.0 kg/m3 has to be drawn.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

(a)

Expert Solution

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.00 to 1.0 kg/m3Ks= 0.03 kg/m3

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.25 kg/m3)(0.03kg/m3)+ (0.25 kg/m3) = 1.34×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(0.50 kg/m3)(0.03kg/m3)+ (0.50 kg/m3) = 1.42×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(0.75 kg/m3)(0.03kg/m3)+ (0.75 kg/m3) = 1.44×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(1.0 kg/m3)(0.03kg/m3)+ (1.0 kg/m3) = 1.46×10−4s−1

S(kg/m3)μ(10−4s−1)0.251.340.501.420.751.441.001.46

By taking all the above data, a graph between μ and S can be plotted like below,

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 16, Problem 16.121P

Figure.1

(b)

Interpretation Introduction

Interpretation:

The density after one hour if the initial S is 0.30 kg/m3 with given initial population density of 5.0×103 cells/m3 has to be determined.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

The first-order reaction kinetics,

ln[A]t= ln[A]0 + k0t

Where,

[A]t= Concentration of reactant at time t[A]0= Initial Concentration of reactantk0= First-order rate constant

(b)

Expert Solution

Answer to Problem 16.121P

The density after one hour has been determined as 8.2×103 cells/m3.

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.30 kg/m3Ks= 0.03 kg/m3t= 1h=3600 s

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.25 kg/m3)(0.03kg/m3)+ (0.25 kg/m3) = 1.363×10−4s−1.

According to the first-order kinetics,

ln[A]t= ln[A]0 + k0tln[A]t= ln[5.0×103] + (1.363×10−4s−1)(3600 s)ln[A]t= 9.00808919[A]t= 8.1689×103 = 8.2×103 cells/m3.

Therefore, the density after one hour is found out to be 8.2×103 cells/m3.

(c)

Interpretation Introduction

Interpretation:

The density after one hour if the initial S is 0.70 kg/m3 has to be determined.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

The first-order reaction kinetics,

ln[A]t= ln[A]0 + k0t

Where,

[A]t= Concentration of reactant at time t[A]0= Initial Concentration of reactantk0= First−order rate constant

(c)

Expert Solution

Answer to Problem 16.121P

The density after one hour if the initial S is 0.70 kg/m3 has been determined as 8.4×103 cells/m3.

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.70 kg/m3Ks= 0.03 kg/m3t= 1h=3600 s

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.70 kg/m3)(0.03kg/m3)+ (0.70 kg/m3) = 1.438356×10−4s−1.

According to the first-order kinetics,

ln[A]t= ln[A]0 + k0tln[A]t= ln[5.0×103] + (1.438356×10−4s−1)(3600 s)ln[A]t= 9.03500135[A]t= 8.39172×103 = 8.4×103 cells/m3.

Therefore, the density after one hour is found out to be 8.4×103 cells/m3.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 16, Problem 16.121P

(a)

Interpretation Introduction

Interpretation:

A Plot of μ vs. S for S between 0.0 and 1.0 kg/m3 has to be drawn.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

(a)

Expert Solution

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.00 to 1.0 kg/m3Ks= 0.03 kg/m3

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.25 kg/m3)(0.03kg/m3)+ (0.25 kg/m3) = 1.34×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(0.50 kg/m3)(0.03kg/m3)+ (0.50 kg/m3) = 1.42×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(0.75 kg/m3)(0.03kg/m3)+ (0.75 kg/m3) = 1.44×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(1.0 kg/m3)(0.03kg/m3)+ (1.0 kg/m3) = 1.46×10−4s−1

S(kg/m3)μ(10−4s−1)0.251.340.501.420.751.441.001.46

By taking all the above data, a graph between μ and S can be plotted like below,

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 16, Problem 16.121P

Figure.1

(b)

Interpretation Introduction

Interpretation:

The density after one hour if the initial S is 0.30 kg/m3 with given initial population density of 5.0×103 cells/m3 has to be determined.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

The first-order reaction kinetics,

ln[A]t= ln[A]0 + k0t

Where,

[A]t= Concentration of reactant at time t[A]0= Initial Concentration of reactantk0= First-order rate constant

(b)

Expert Solution

Answer to Problem 16.121P

The density after one hour has been determined as 8.2×103 cells/m3.

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.30 kg/m3Ks= 0.03 kg/m3t= 1h=3600 s

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.25 kg/m3)(0.03kg/m3)+ (0.25 kg/m3) = 1.363×10−4s−1.

According to the first-order kinetics,

ln[A]t= ln[A]0 + k0tln[A]t= ln[5.0×103] + (1.363×10−4s−1)(3600 s)ln[A]t= 9.00808919[A]t= 8.1689×103 = 8.2×103 cells/m3.

Therefore, the density after one hour is found out to be 8.2×103 cells/m3.

(c)

Interpretation Introduction

Interpretation:

The density after one hour if the initial S is 0.70 kg/m3 has to be determined.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

The first-order reaction kinetics,

ln[A]t= ln[A]0 + k0t

Where,

[A]t= Concentration of reactant at time t[A]0= Initial Concentration of reactantk0= First−order rate constant

(c)

Expert Solution

Answer to Problem 16.121P

The density after one hour if the initial S is 0.70 kg/m3 has been determined as 8.4×103 cells/m3.

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.70 kg/m3Ks= 0.03 kg/m3t= 1h=3600 s

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.70 kg/m3)(0.03kg/m3)+ (0.70 kg/m3) = 1.438356×10−4s−1.

According to the first-order kinetics,

ln[A]t= ln[A]0 + k0tln[A]t= ln[5.0×103] + (1.438356×10−4s−1)(3600 s)ln[A]t= 9.03500135[A]t= 8.39172×103 = 8.4×103 cells/m3.

Therefore, the density after one hour is found out to be 8.4×103 cells/m3.

Answer 6

Chapter 16, Problem 16.121P

(a)

Interpretation Introduction

Interpretation:

A Plot of μ vs. S for S between 0.0 and 1.0 kg/m3 has to be drawn.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

(a)

Expert Solution

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.00 to 1.0 kg/m3Ks= 0.03 kg/m3

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.25 kg/m3)(0.03kg/m3)+ (0.25 kg/m3) = 1.34×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(0.50 kg/m3)(0.03kg/m3)+ (0.50 kg/m3) = 1.42×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(0.75 kg/m3)(0.03kg/m3)+ (0.75 kg/m3) = 1.44×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(1.0 kg/m3)(0.03kg/m3)+ (1.0 kg/m3) = 1.46×10−4s−1

S(kg/m3)μ(10−4s−1)0.251.340.501.420.751.441.001.46

By taking all the above data, a graph between μ and S can be plotted like below,

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 16, Problem 16.121P

Figure.1

Answer 7

Chapter 16, Problem 16.121P

(a)

Interpretation Introduction

Interpretation:

A Plot of μ vs. S for S between 0.0 and 1.0 kg/m3 has to be drawn.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

Answer 8

(a)

Expert Solution

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.00 to 1.0 kg/m3Ks= 0.03 kg/m3

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.25 kg/m3)(0.03kg/m3)+ (0.25 kg/m3) = 1.34×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(0.50 kg/m3)(0.03kg/m3)+ (0.50 kg/m3) = 1.42×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(0.75 kg/m3)(0.03kg/m3)+ (0.75 kg/m3) = 1.44×10−4s−1μ = μmaxSKs+ S = (1.5×10−4s−1)(1.0 kg/m3)(0.03kg/m3)+ (1.0 kg/m3) = 1.46×10−4s−1

S(kg/m3)μ(10−4s−1)0.251.340.501.420.751.441.001.46

By taking all the above data, a graph between μ and S can be plotted like below,

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 16, Problem 16.121P

Figure.1

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The density after one hour if the initial S is 0.30 kg/m3 with given initial population density of 5.0×103 cells/m3 has to be determined.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

The first-order reaction kinetics,

ln[A]t= ln[A]0 + k0t

Where,

[A]t= Concentration of reactant at time t[A]0= Initial Concentration of reactantk0= First-order rate constant

(b)

Expert Solution

Answer to Problem 16.121P

The density after one hour has been determined as 8.2×103 cells/m3.

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.30 kg/m3Ks= 0.03 kg/m3t= 1h=3600 s

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.25 kg/m3)(0.03kg/m3)+ (0.25 kg/m3) = 1.363×10−4s−1.

According to the first-order kinetics,

ln[A]t= ln[A]0 + k0tln[A]t= ln[5.0×103] + (1.363×10−4s−1)(3600 s)ln[A]t= 9.00808919[A]t= 8.1689×103 = 8.2×103 cells/m3.

Therefore, the density after one hour is found out to be 8.2×103 cells/m3.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The density after one hour if the initial S is 0.30 kg/m3 with given initial population density of 5.0×103 cells/m3 has to be determined.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

The first-order reaction kinetics,

ln[A]t= ln[A]0 + k0t

Where,

[A]t= Concentration of reactant at time t[A]0= Initial Concentration of reactantk0= First-order rate constant

Answer 14

(b)

Expert Solution

Answer to Problem 16.121P

The density after one hour has been determined as 8.2×103 cells/m3.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.30 kg/m3Ks= 0.03 kg/m3t= 1h=3600 s

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.25 kg/m3)(0.03kg/m3)+ (0.25 kg/m3) = 1.363×10−4s−1.

According to the first-order kinetics,

ln[A]t= ln[A]0 + k0tln[A]t= ln[5.0×103] + (1.363×10−4s−1)(3600 s)ln[A]t= 9.00808919[A]t= 8.1689×103 = 8.2×103 cells/m3.

Therefore, the density after one hour is found out to be 8.2×103 cells/m3.

Answer 19

(c)

Interpretation Introduction

Interpretation:

The density after one hour if the initial S is 0.70 kg/m3 has to be determined.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

The first-order reaction kinetics,

ln[A]t= ln[A]0 + k0t

Where,

[A]t= Concentration of reactant at time t[A]0= Initial Concentration of reactantk0= First−order rate constant

(c)

Expert Solution

Answer to Problem 16.121P

The density after one hour if the initial S is 0.70 kg/m3 has been determined as 8.4×103 cells/m3.

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.70 kg/m3Ks= 0.03 kg/m3t= 1h=3600 s

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.70 kg/m3)(0.03kg/m3)+ (0.70 kg/m3) = 1.438356×10−4s−1.

According to the first-order kinetics,

ln[A]t= ln[A]0 + k0tln[A]t= ln[5.0×103] + (1.438356×10−4s−1)(3600 s)ln[A]t= 9.03500135[A]t= 8.39172×103 = 8.4×103 cells/m3.

Therefore, the density after one hour is found out to be 8.4×103 cells/m3.

Answer 20

(c)

Interpretation Introduction

Interpretation:

The density after one hour if the initial S is 0.70 kg/m3 has to be determined.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

μ = μmaxSKs+ S

Where,

μ = first-order growth rate (s-1)μmax= Maximum growth rateS = Substrate concentrationKs= the value of S that gives one-half of the maximum growth rate(in kg/m3)

The first-order reaction kinetics,

ln[A]t= ln[A]0 + k0t

Where,

[A]t= Concentration of reactant at time t[A]0= Initial Concentration of reactantk0= First−order rate constant

Answer 21

(c)

Expert Solution

Answer to Problem 16.121P

The density after one hour if the initial S is 0.70 kg/m3 has been determined as 8.4×103 cells/m3.

Answer 22

(c)

Expert Solution

Answer 23

(c)

Expert Solution

Answer 24

Expert Solution

Answer 25

Explanation of Solution

Given data:

μmax= 1.5×10−4s−1S = 0.70 kg/m3Ks= 0.03 kg/m3t= 1h=3600 s

Calculation of first-order growth rate:

μ = μmaxSKs+ S = (1.5×10−4s−1)(0.70 kg/m3)(0.03kg/m3)+ (0.70 kg/m3) = 1.438356×10−4s−1.

According to the first-order kinetics,

ln[A]t= ln[A]0 + k0tln[A]t= ln[5.0×103] + (1.438356×10−4s−1)(3600 s)ln[A]t= 9.03500135[A]t= 8.39172×103 = 8.4×103 cells/m3.

Therefore, the density after one hour is found out to be 8.4×103 cells/m3.

Answer 26

Ch. 16.2 - Balance the following equation and express the...Ch. 16.2 - Prob. 16.1BFP Ch. 16.3 - Prob. 16.2AFP Ch. 16.3 - Prob. 16.2BFP Ch. 16.3 - Find the rate law, the individual and overall...Ch. 16.3 - For the reaction at 0°C, the following data were...Ch. 16.3 - Prob. 16.4AFP Ch. 16.3 - Prob. 16.4BFP Ch. 16.4 - At 25°C, hydrogen iodide breaks down very slowly...Ch. 16.4 - Prob. 16.5BFP

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