Answer:
"0.60 g" is the appropriate solution.
Explanation:
The given values are:
Volume of base,
= 30 ml
Molarity of base,
= 0.05 m
Molar mass of acid,
= 400 g/mol
As we know,
⇒
On substituting the values, we get
⇒
⇒
⇒
hence,
⇒
On substituting the values, we get
⇒
⇒
⇒
Answer:
a. The second run will be faster.
d. The second run has twice the surface area.
Explanation:
The rate of a reaction is proportional to the surface area of a catalyst. Given the volume (V) of a sphere, we can find its surface area (A) using the following expression.
The area of the 10.0 cm³-sphere is:
The area of each 1.25 cm³-sphere is:
The total area of the 8 1.25cm³-spheres is 8 × 5.61 cm² = 44.9 cm²
The ratio of 8 1.25cm³-sphere to 10.0 cm³-sphere is 44.9 cm²/22.4 cm² = 2.00
Since the surface area is doubled, the second run will be faster.
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