the chloride of a metal M contains 47.25% of metal 1.0 gram of metal would be displaced from a compound by 0.88 gram of another
1 answer:
Answer:
The equivalent weight of M is approximately 31.8 g
The equivalent weight of N is approximately 27.98 g
Explanation:
The given parameters are;
The percentage of the the metal M in in the chloride = 47.25%
Where by the chemical formula for the metal chloride is MClₓ, we have;
47.25% of the mass of MClₓ = Mass of M = W
Therefore, we have;
0.4725 × (W + 35.5·x) = W
0.4725·W + 0.4725×35.5×x = W
W - 0.4725·W = 16.77·x
0.5275·W = 16.77·x
W/x = 16.77/0.5275 = 31.799 = The equivalent weight of M
The equivalent weight of M = 31.799 ≈ 31.8 g
Given that 1 gram of M is displaced by 0.88 gram of N, then the equivalent weight of N that will displace 31.799 = 0.88 × 31.799 ≈ 27.98 g
The equivalent weight of N = 27.98 g.
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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