Answer: to calculate pH use -log[H+] or - log[OH-]..the solution is basic as the “NaOH” is attached to a hydroxide.Since we need to find the pH (per hydrogen) and not the pOH( per hydroxide) we need to find the pOH of the substance first then we subtract that by 14 so we can arrive at the pH of the substance.
Explanation: So -log( 1 x 10^(-5)) = 5 which is the pOH.Now we subtract that by 14 which gives us -9 and now you’d multiply that by -1 bcuz we can’t have a negative so the pH of the substance is 9
Answer: 
of NaOH
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.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
Putting in the values:
![10.020=-\log[H^+]](https://tex.z-dn.net/?f=10.020%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=9.55\times 10^{-11}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D9.55%5Ctimes%2010%5E%7B-11%7D)
![[H^+][OH^-]=10^{-14}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%5BOH%5E-%5D%3D10%5E%7B-14%7D)
![[OH^-]=\frac{10^{-14}}{9.55\times 10^{-11}}=1.05\times 10^{-4}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B9.55%5Ctimes%2010%5E%7B-11%7D%7D%3D1.05%5Ctimes%2010%5E%7B-4%7DM)
moles = 
Mass of 
Thus of NaOH is needed to prepare 463 mL of solution with a pH of 10.020
Answer:
chlorine, fluorine, bromine, iodine, Xenon, and radon
Answer:
yes the answer is true because heat changes duriing epalthy change
