After finding the oxidation states of atoms, you identify the half reactions (option c).
The half reactions are given by the change of the oxidation states of the atoms.
For example if Cu is in the left side with oxidation state 0 and in the other side with oxidation state 2+, then there you have a half reaction (oxidation reaction). And if you have O with oxidation state 0 in the left side and with oxidation state 2- in the right side, there you have other half reaction (reducing reaction).
Answer:
136
Explanation:
The Mass Number is the combination of the amount of Protons and Neutrons in an element, so if the total mass is 222, and the amount of protons is 86, then you can do 86 + x = 222 to find that x is equal to 136
Answer:
HCl(aq) + KOH(aq) --> KCl(aq) + H2O(l)
Explanation:
A neutralization reaction is the process between an acid and a base (there are a number of different ways to define acids and bases). An acid is a compound, which dissolves in water by releasing H+ ions, and a base is a compound, which dissolves in water by releasing OH- ions (by Arrhenius' definition, the simplest one). In this case, the neutralization reaction is the process between HCl (hydrochloric acid) - an acid, and KOH (potassium hydroxide) - a base.
If the equation is complete the products would be manganese chloride and oxygen gas would be given off.
MnCl2 + O2
Hello!
First, we need to determine the pKa of the base. It can be found applying the following equation:
Now, we can apply the
Henderson-Hasselbach's equation in the following way:
![pH=pKa+log( \frac{[CH_3NH_2]}{[CH_3NH_3Cl]} )=10,65+log( \frac{0,18M}{0,73M} )=10,04](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%20%5Cfrac%7B%5BCH_3NH_2%5D%7D%7B%5BCH_3NH_3Cl%5D%7D%20%29%3D10%2C65%2Blog%28%20%5Cfrac%7B0%2C18M%7D%7B0%2C73M%7D%20%29%3D10%2C04)
So,
the pH of this buffer solution is 10,04Have a nice day!
