Answer: pH = 2,897 , basic![[H+][OH-] = 10^{-14} ==> [H+] = \frac{10^{-14}}{7,89*10^{-12} } =\frac{1}{789} \\pH= -lg([H+]) = 2,897 \\pH basic](https://tex.z-dn.net/?f=%5BH%2B%5D%5BOH-%5D%20%3D%2010%5E%7B-14%7D%20%3D%3D%3E%20%5BH%2B%5D%20%3D%20%5Cfrac%7B10%5E%7B-14%7D%7D%7B7%2C89%2A10%5E%7B-12%7D%20%7D%20%3D%5Cfrac%7B1%7D%7B789%7D%20%5C%5CpH%3D%20-lg%28%5BH%2B%5D%29%20%3D%202%2C897%20%5C%5CpH%3C7%20%3D%3D%3E%20basic)
Explanation:
What statement?
Light moves faster through a vacuum because no particles are existent to absorb it.
The electron geometry is tetrahedral and the molecular geometry is tetrahedral. If a molecule of CH3OCH3 is to be drawn, the two carbons would have four single bonds and the middle O would have two single bonds and the two lone pairs. Molecular geometry does not consider the lone pairs as bonds like in electron domain geometry. However, since the carbons do not contain any lone pairs, its electrons domain and molecular geometry will be the same. Therefore; Both carbons are tetrahedral for electron domain geometry and molecular geometries. The O is tetrahedral for the electron domain geometry and bent for molecular geometry.
Answer:
1.135 M.
Explanation:
- For the reaction: <em>2HI → H₂ + I₂,</em>
The reaction is a second order reaction of HI,so the rate law of the reaction is: Rate = k[HI]².
- To solve this problem, we can use the integral law of second-order reactions:
<em>1/[A] = kt + 1/[A₀],</em>
where, k is the reate constant of the reaction (k = 1.57 x 10⁻⁵ M⁻¹s⁻¹),
t is the time of the reaction (t = 8 hours x 60 x 60 = 28800 s),
[A₀] is the initial concentration of HI ([A₀] = ?? M).
[A] is the remaining concentration of HI after hours ([A₀] = 0.75 M).
∵ 1/[A] = kt + 1/[A₀],
∴ 1/[A₀] = 1/[A] - kt
∴ 1/[A₀] = [1/(0.75 M)] - (1.57 x 10⁻⁵ M⁻¹s⁻¹)(28800 s) = 1.333 M⁻¹ - 0.4522 M⁻¹ = 0.8808 M⁻¹.
∴ [A₀] = 1/(0.0.8808 M⁻¹) = 1.135 M.
<em>So, the concentration of HI 8 hours earlier = 1.135 M.</em>
