Answer:
The answer to the question is
The rate constant for the reaction is 1.056×10⁻³ M/s
Explanation:
To solve the question, e note that
For a zero order reaction, the rate law is given by
[A] = -k×t + [A]₀
This can be represented by the linear equation y = mx + c
Such that y = [A], m which is the gradient is = -k, and the intercept c = [A]₀
Therefore the rate constant k which is the gradient is given by
Gradient = ![\frac{[A]_{2} - [A]_{1} }{t_{2} - t_{1} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5D_%7B2%7D%20-%20%5BA%5D_%7B1%7D%20%20%7D%7Bt_%7B2%7D%20-%20t_%7B1%7D%20%20%7D)
where [A]₁ = 8.10×10⁻² M and [A]₂ = 1.80×10⁻³ M
= 
= -0.001056 M/s = -1.056×10⁻³ M/s
Threfore k = 1.056×10⁻³ M/s
Answer:
Mass of one electron is 9.1 × 10⁻³¹ kg
Mass of one proton is 1.673 × 10⁻²⁷ Kg
Mass of one neutron is 1.675 × 10⁻²⁷ Kg
<u>-TheUnknownScientist</u><u> 72</u>
There would be 67 left because you do
Q1)
We have been given the OH⁻ concentration, therefore we first need to find the pOH value and then the pH value.
pOH = -log [OH⁻]
pOH = -log (0.225 M)
pOH = 0.65
pH + pOH = 14
pH = 14 - 0.65 = 13.35
Q2)
pOH = -log[OH⁻]
pOH = -log (0.0015 M)
pOH = 2.82
pH + pOH = 14
pH = 14 - 2.82
pH = 11.18
Answer:
Correct option is B)
According to de-Broglie,
λ=mvh=6.023×10232×5×104cm/sec6.62×10−27ergsec=4×10−8cm=4Ao
