3. Make an argument for placing hydrogen in the halogen group rather than the alkali metals. 4. Compare and contrast ionization
1 answer:
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Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,
pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
Now put all the given values in this expression, we get:
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100
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.
Answer:
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