Answer:
take these as notes ig:)
Explanation:
reaction equations to make connections between the reactants and products in reactions. ... pathway: mol of O2 → mol of H2O → grams of H2O. 3.5 mol ... 25 g KClO3 ... How many grams of hydrogen gas are produced from the reaction of 221.
The way I would explain it is quite difficult to understand, so this is what Google says. "The wavelength (or equivalently, frequency) of the photon is determined by the difference in energy between the two states. These emitted photons form the element's spectrum. The fact that only certain colors appear in an element's atomic emission spectrum means that only certain frequencies of light are emitted." I hope this helped.
Answer: Solution A : ![[H_3O^+]=0.300\times 10^{-7}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D0.300%5Ctimes%2010%5E%7B-7%7DM)
Solution B : ![[OH^-]=0.107\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.107%5Ctimes%2010%5E%7B-5%7DM)
Solution C : ![[OH^-]=0.177\times 10^{-10}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.177%5Ctimes%2010%5E%7B-10%7DM)
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

![[H_3O^+][OH^-]=10^{-14}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%5BOH%5E-%5D%3D10%5E%7B-14%7D)
a. Solution A: ![[OH^-]=3.33\times 10^{-7}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D3.33%5Ctimes%2010%5E%7B-7%7DM)
![[H_3O^+]=\frac{10^{-14}}{3.33\times 10^{-7}}=0.300\times 10^{-7}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B3.33%5Ctimes%2010%5E%7B-7%7D%7D%3D0.300%5Ctimes%2010%5E%7B-7%7DM)
b. Solution B : ![[H_3O^+]=9.33\times 10^{-9}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D9.33%5Ctimes%2010%5E%7B-9%7DM)
![[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B9.33%5Ctimes%2010%5E%7B-9%7D%7D%3D0.107%5Ctimes%2010%5E%7B-5%7DM)
c. Solution C : ![[H_3O^+]=5.65\times 10^{-4}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D5.65%5Ctimes%2010%5E%7B-4%7DM)
![[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B5.65%5Ctimes%2010%5E%7B-4%7D%7D%3D0.177%5Ctimes%2010%5E%7B-10%7DM)