<u>Answer:</u>
<em>A. 10.25</em>
<em></em>
<u>Explanation:</u>
Pkb =4.77
So pka = 14 - pka = 9.23
Initial 0.50M 0 0
Change -x +x +x
Equilibrium 0.50M-x +x +x
(-x is neglected) so we get
![pH=-log[H^3 O^+]\\\\pH=-log[1.72\times10^{-5}]\\\\pH=4.76](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E3%20O%5E%2B%5D%5C%5C%5C%5CpH%3D-log%5B1.72%5Ctimes10%5E%7B-5%7D%5D%5C%5C%5C%5CpH%3D4.76)
pOH = 14 - pH
= 14 - 4.76
pOH = 9.24 is the answer
Option A - 10.25 is the answer which is close to 9.24
B producers. Think of it as if they’re producing the goods
The balanced equation for the reaction is as follows
2H₂ + O₂ --> 2H₂O
stoichiometry of H₂ to O₂ is 2:1
number of H₂ moles - 30.0 g / 2 g/mol = 15 mol
number of O₂ moles - 80.0 g / 32 g/mol = 2.5 mol
limiting reactant is the reagent in which only a fraction is used up in the reaction
if H₂ is the limiting reactant
if 2 mol of H₂ requires 1 mol of O₂
then 15 mol of H₂ requires 1/2 x 15.0 = 7.5 mol of O₂
but only 2.5 mol of O₂ is required
this means that O₂ is the limiting reagentt and H₂ is in excess
Answer:
monoxide
dioxide
trioxide
tetroxide
pentoxide
hexoxide
heptoxide
octoxide
nonoxide
Explanation:
Just the way it is. If there is an -a or -o infront of the oxide(or another substance that starts with vowel), the a is often dropped.
