Hemoglobin molecules in blood bind oxygen and carry it to cells, where it takes part in metabolism. The binding of oxygen hemoglobin(aq) + O2(aq) 88n hemoglobin O2(aq) is rst order in hemoglobin and rst order in dissolved oxygen, with a rate constant of 4 × 107 L mol 1 s 1. Calcu- late the initial rate at which oxygen will be bound to hemo- globin if the concentration of hemoglobin is 2 × 10 9 M and that of oxygen is 5 × 10 5 M.

Without wasting much of our time, Here is the correct question.

Hemoglobin molecules in blood bind oxygen and carry it to cells, where it takes part in metabolism. The binding of oxygen hemoglobin(aq) + O2(aq) -------> hemoglobin O2(aq) is first order in hemoglobin and first order in dissolved oxygen, with a rate constant of 4 × 10⁷ L mol⁻¹ s⁻¹. Calculate the initial rate at which oxygen will be bound to hemoglobin if the concentration of hemoglobin is 2 × 10⁻⁹ M and that of oxygen is 5 × 10⁻⁵M.

4 × 10⁻⁶ M s⁻¹

Explanation:

The equation for the reaction between Hemoglobin molecules in blood that binds with oxygen molecule can be represent by:

 +     --------->

Now, we are also being told to calculate only!, the  initial rate at which oxygen will be bound to hemoglobin.

So, If it is first order in hemoglobin and also first order in Oxygen molecule at the initial rate of the the reaction, therefore, the rate  for the reaction can be expressed as :

rate = k

Let's not forget that we are so given some parameters;

where

k (rate constant) = 4 × 10⁷ L mol⁻¹ s⁻¹

[ ] = 2 × 10⁻⁹ M

[    ]  =  5 × 10⁻⁵ M

Substituting our data given into the above rate formula, we have:

rate = (4 × 10⁷ L mol⁻¹ s⁻¹) × (2 × 10⁻⁹ M) × (5 × 10⁻⁵ M)

rate = 4 × 10⁻⁶ M s⁻¹     ( given that 1 M = 1 mol L⁻¹ )

∴ the initial rate at which oxygen will be bound to hemoglobin = 4 × 10⁻⁶ M s⁻¹

The fraction that this is true for = 7/13

Step-by-step explanation:

From the above question

Let the numerator be represented by a

Let the denominator be represented by b

If you add 5 to both the numerator and the denominator of a fraction, you obtain 2/3

This means:

a + 5/b + 5 = 2/3

Cross Multiply

3(a + 5) = 2(b + 5)

3a + 15 = 2b + 10

Collect like terms

3a - 2b = 10 - 15

3a - 2b = -5Equation 1

If you subtract 5 from both the numerator and the denominator of the same fraction, you obtain 1/4

This means:

a - 5/b - 5 = 1/4

Cross Multiply

4(a - 5) = 1(b - 5)

4a - 20 = b - 5

Collect like terms

4a - b = 20 - 5

4a - b = 15Equation 2

b = 4a - 15

3a - 2b = -5Equation 1

4a - b = 15Equation 2

Substitute 4a - 15 for b in equation 1

3a - 2b = -5Equation 1

3a - 2(4a - 15) = -5

3a - 8a + 30 = -5

Collect like terms

3a - 8a = -5 - 30

-5a = -35

a = -35/-5

a = 7

Therefore, the numerator of the fraction = 7

Substitute 7 for a in Equation 2

4a - b = 15Equation 2

4 × 7 - b = 15

28 - b =15

28 - 15 = b

b = 13

The denominator = b is 13.

Therefore,the fraction which this is true for = 7/13

To confirm

a) If you add 5 to both the numerator and the denominator of a fraction, you obtain 2/3

This means:

a + 5/b + 5 = 2/3

7 + 5/ 13 + 5 = 2/3

12/18 = 2/3

Divide numerator and denominator by of the left hand side by 6

12÷ 6/ 18 ÷ 6 = 2/3

2/3 =2/3

If you subtract 5 from both the numerator and the denominator of the same fraction, you obtain 1/4

This means:

a - 5/b - 5 = 1/4

7 - 5/13 - 5 = 1/4

2/8 = 1/4

Divide the numerator and denominator of the left hand side by 2

2÷2/8 ÷ 2 = 1/4

1/4 = 1/4

From the above confirmation, the fraction that this is true for is 7/13