How many miles of NaCl are in 250 mL of a 0.4 M solution?
1 answer:
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Answer:
a) No molecules of hydrogen
b) four molecules of ammonia
c) four left molecules of nitrogen.
Explanation:
The balanced reaction between nitrogen and hydrogen molecules to give ammonia molecules is:
Thus one molecule of nitrogen will react with three molecules of hydrogen to give two molecules of ammonia.
We have six molecules of each nitrogen and hydrogen in the closed container and they undergo complete reaction it means the limiting reagent is hydrogen. For six molecules of nitrogen, eighteen molecules of hydrogen will be required.
So six molecules of hydrogen will react with two molecules of nitrogen to give four molecules of ammonia.
The product mixture will have
a) No molecules of hydrogen
b) four molecules of ammonia
c) four left molecules of nitrogen.
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