Answer:
the last one is c and the other ones are aa
Explanation:
Answer:
The pH of the 0.100 M solution of hydobromous acid HBrO is 4.843
Explanation:
Here we have the reaction given as follows;
HBrO ⇄ H⁺ + BrO⁻¹
Therefore;
HBrO ⇄ H⁺ + BrO⁻¹
Initial concentration of HBrO = 0.100 M decomposes partly to form x moles of each of H⁺ and BrO⁻¹. That is
HBrO ⇄ H⁺ + BrO⁻¹
Initial 0.1 0 0
Change -x +x +x
Equilibrium 0.1 - x x x
Hence;
![Ka = \frac{[H^+][BrO^{-1}]}{[HBrO]} =\frac{x \times x }{0.1 - x} = \frac{x^2}{0.1 -x} = 2.06 \times 10^{-9}](https://tex.z-dn.net/?f=Ka%20%3D%20%20%5Cfrac%7B%5BH%5E%2B%5D%5BBrO%5E%7B-1%7D%5D%7D%7B%5BHBrO%5D%7D%20%3D%5Cfrac%7Bx%20%5Ctimes%20x%20%7D%7B0.1%20-%20x%7D%20%3D%20%5Cfrac%7Bx%5E2%7D%7B0.1%20-x%7D%20%3D%202.06%20%5Ctimes%2010%5E%7B-9%7D)
(0.1 - x) × 2.06×10⁻⁹ = x²
x² + 2.06×10⁻⁹·x - 0.1 = 0
Factorizing gives;
(x + 1.4354×10⁻⁵)(x - 1.4352×10⁻⁵) = 0
Therefore, x = 1.4352×10⁻⁵ M or -1.4354×10⁻⁵ M
We take the positive value as x is the concentration of the ions in the solution;
From the above [H⁺] = 1.4352×10⁻⁵ M
pH = -log[H⁺] = -log(1.4352×10⁻⁵) = 4.843
The pH of the 0.100 M solution of hydobromous acid HBrO at 298 K where the Ka = 2.06×10⁻⁹ = 4.843.
Answer: pH is a measure of how acidic/basic water is. The range goes from 0 to 14, with 7 being neutral. pHs of less than 7 indicate acidity, whereas a pH of greater than 7 indicates a base. pH is really a measure of the relative amount of free hydrogen and hydroxyl ions in the water.
Explanation:
look it up on the internet
Answer:

Explanation:
By definition one <em>half-life</em> is the time to reduce the initial concentration to half.
For a <em>second order reaction </em>the rate law equations are:
![\dfrac{d[B]}{dt}=-k[B]^2](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5BB%5D%7D%7Bdt%7D%3D-k%5BB%5D%5E2)
![\dfrac{1}{[B]}=\dfrac{1}{[B]_0}+kt](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5BB%5D%7D%3D%5Cdfrac%7B1%7D%7B%5BB%5D_0%7D%2Bkt)
The <em>half-life</em> equation is:
![t_{1/2}=\dfrac{1}{k[A]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cdfrac%7B1%7D%7Bk%5BA%5D_0%7D)
Thus, substitute the<em> rate constant</em>
and the <em>half-life </em>time <em>224s</em> to find [A]₀:
![224s=\dfrac{1}{1.30\times10^{-3}M^{-1}\cdot s^{-1}[A]_0}](https://tex.z-dn.net/?f=224s%3D%5Cdfrac%7B1%7D%7B1.30%5Ctimes10%5E%7B-3%7DM%5E%7B-1%7D%5Ccdot%20s%5E%7B-1%7D%5BA%5D_0%7D)
![[A]_o=0.291M](https://tex.z-dn.net/?f=%5BA%5D_o%3D0.291M)