Answer:
Explanation:
In this, we can with the <u>ionization equation</u> for the hydrazoic acid (). So:
Now, due to the Ka constant value, we have to use the whole equilibrium because this <u>is not a strong acid</u>. So, we have to write the <u>Ka expression</u>:
For each mol of produced we will have 1 mol of . So, we can use <u>"X" for the unknown</u> values and replace in the Ka equation:
Additionally, we have to keep in mind that is a reagent, this means that we will be <u>consumed</u>. We dont know how much acid would be consumed but we can express a<u> subtraction from the initial value</u>, so:
Finally, we can put the ka value and <u>solve for "X"</u>:
So, we have a concentration of 0.000285 for . With this in mind, we can calculate the <u>pH value</u>:
I hope it helps!
the second options
the number of protons and electrons found in that element
Answer:
So, when bromine comes near the ethylene pi bond, it <u><em>attacks the electron and grabs it</em></u>, but the second bromine steals it away and departs as Br-. After all, bromine is more electronegative than carbon; the first bromine is just a link in the chain of events that allows the second bromine to escape with an electron.
the configuration for chlorine is [Ne] 3s2 3p5
Answer:
The formula for Lead(ll) Oxide is PbO.
Explanation:
The symbol for Lead is Pb.
The symbol for Oxygen is O.
Given,Oxidation state of Pb = +2
We know,
Oxidation state of Oxygen in all Oxides is the same and is equal to -2,
Let the no. of Pb atoms in the oxide = x;
no.of Oxygen atoms = y;
Concept:
Sum of oxidation states of all atoms in any compound is equal to zero.
Applying concept,we get,
x(+2) + y(-2) = 0
x(+2) = y(+2)
Therefore, x = y
and, x : y = 1:1 ;
Therefore the empirical and the correct formula of the oxide is ,which is .