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<em>I'm</em><em> </em><em>Abraar </em><em>Shaikh</em>
here is your explanation
Explanation:
People with high IQ might have bad grades. It is not necessary that high IQ requires higher grades. Mind of a person with high IQ works at a faster rate so they tend to ignore normal problems almost everywhere even in the examinations.
Answer:
rate = k[A][B] where k = k₂K
Explanation:
Your mechanism is a slow step with a prior equilibrium:
![\begin{array}{rrcl}\text{Step 1}:& \text{A + B} & \xrightarrow [k_{-1}]{k_{1}} & \text{C}\\\text{Step 2}: & \text{C + A} & \xrightarrow [ ]{k_{2}} & \text{D}\\\text{Overall}: & \text{2A + B} & \longrightarrow \, & \text{D}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brrcl%7D%5Ctext%7BStep%201%7D%3A%26%20%5Ctext%7BA%20%2B%20B%7D%20%26%20%5Cxrightarrow%20%5Bk_%7B-1%7D%5D%7Bk_%7B1%7D%7D%20%26%20%5Ctext%7BC%7D%5C%5C%5Ctext%7BStep%202%7D%3A%20%26%20%5Ctext%7BC%20%2B%20A%7D%20%26%20%5Cxrightarrow%20%5B%20%5D%7Bk_%7B2%7D%7D%20%26%20%5Ctext%7BD%7D%5C%5C%5Ctext%7BOverall%7D%3A%20%26%20%5Ctext%7B2A%20%2B%20B%7D%20%26%20%5Clongrightarrow%20%5C%2C%20%26%20%5Ctext%7BD%7D%5C%5C%5Cend%7Barray%7D)
(The arrow in Step 1 should be equilibrium arrows).
1. Write the rate equations:
![-\dfrac{\text{d[A]}}{\text{d}t} = -\dfrac{\text{d[B]}}{\text{d}t} = -k_{1}[\text{A}][\text{B}] + k_{1}[\text{C}]\\\\\dfrac{\text{d[C]}}{\text{d}t} = k_{1}[\text{A}][\text{B}] - k_{2}[\text{C}]\\\\\dfrac{\text{d[D]}}{\text{d}t} = k_{2}[\text{C}]](https://tex.z-dn.net/?f=-%5Cdfrac%7B%5Ctext%7Bd%5BA%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20-%5Cdfrac%7B%5Ctext%7Bd%5BB%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20-k_%7B1%7D%5B%5Ctext%7BA%7D%5D%5B%5Ctext%7BB%7D%5D%20%2B%20k_%7B1%7D%5B%5Ctext%7BC%7D%5D%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7Bd%5BC%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20k_%7B1%7D%5B%5Ctext%7BA%7D%5D%5B%5Ctext%7BB%7D%5D%20-%20k_%7B2%7D%5B%5Ctext%7BC%7D%5D%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7Bd%5BD%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20k_%7B2%7D%5B%5Ctext%7BC%7D%5D)
2. Derive the rate law
Assume k₋₁ ≫ k₂.
Then, in effect, we have an equilibrium that is only slightly disturbed by C slowly reacting to form D.
In an equilibrium, the forward and reverse rates are equal:
k₁[A][B] = k₋₁[C]
[C] = (k₁/k₋₁)[A][B] = K[A][B] (K is the equilibrium constant)
rate = d[D]/dt = k₂[C] = k₂K[A][B] = k[A][B]
The rate law is
rate = k[A][B] where k = k₂K
Answer:
0.51M
Explanation:
Given parameters:
Initial volume of NaBr = 340mL
Initial molarity = 1.5M
Final volume = 1000mL
Unknown:
Final molarity = ?
Solution;
This is a dilution problem whereas the concentration of a compound changes from one to another.
In this kind of problem, we must establish that the number of moles still remains the same.
number of moles initially before diluting = number of moles after dilution
Number of moles = Molarity x volume
Let us find the number of moles;
Number of moles = initial volume x initial molarity
Convert mL to dm³;
1000mL = 1dm³
340mL gives
= 0.34dm³
Number of moles = initial volume x initial molarity = 0.34 x 1.5 = 0.51moles
Now to find the new molarity/concentration;
Final molarity =
=
= 0.51M
We can see a massive drop in molarity this is due to dilution of the initial concentration.
Answer:
20.95 g of caffeine, C₈H₁₀N₄O₂
Explanation:
From the question given above, the following data were obtained:
Number of molecules of C₈H₁₀N₄O₂ = 6.5×10²² molecules
Mass of C₈H₁₀N₄O₂ =?
From Avogadro's hypothesis,
1 mole of C₈H₁₀N₄O₂ = 6.02×10²³ molecules
Next, we shall determine the mass of 1 mole of C₈H₁₀N₄O₂. This can be obtained as follow:
1 mole of C₈H₁₀N₄O₂ = (8×12) + (10×1) + (4×14) + (2×16)
= 96 + 10 + 56 + 32
1 mole of C₈H₁₀N₄O₂ = 194 g
Thus,
194 g of C₈H₁₀N₄O₂ = 6.02×10²³ molecules
Finally, we shall determine the mass of caffeine, C₈H₁₀N₄O₂ that contains 6.5×10²² molecules. This can be obtained as follow:
6.02×10²³ molecules = 194 g of C₈H₁₀N₄O₂
Therefore,
6.5×10²² molecules = (6.5×10²² × 194) / 6.02×10²³
6.5×10²² molecules = 20.95 g of C₈H₁₀N₄O₂.
Therefore, 20.95 g of caffeine, C₈H₁₀N₄O₂ contains 6.5×10²² molecules