The pH of the buffer is 6.1236.
Explanation:
The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.
![pKa=-log[H] = - log [ 5.66 * 10^{-7}]\\ \\pka = 7 - log (5.66)=7-0.753=6.247\\\\pka = 6.247](https://tex.z-dn.net/?f=pKa%3D-log%5BH%5D%20%3D%20-%20log%20%5B%205.66%20%2A%2010%5E%7B-7%7D%5D%5C%5C%20%5C%5Cpka%20%3D%207%20-%20log%20%285.66%29%3D7-0.753%3D6.247%5C%5C%5C%5Cpka%20%3D%206.247)
The pH of the buffer can be known as
![pH = pK_{a} + log[\frac{[A-]}{[HA]}}]](https://tex.z-dn.net/?f=pH%20%3D%20pK_%7Ba%7D%20%2B%20log%5B%5Cfrac%7B%5BA-%5D%7D%7B%5BHA%5D%7D%7D%5D)
The concentration of ![[A^{-}] = Moles of [A]/Total volume = 0.608/2 = 0.304 M\\](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%5D%20%3D%20Moles%20of%20%5BA%5D%2FTotal%20volume%20%3D%200.608%2F2%20%3D%200.304%20M%5C%5C)
Similarly, the concentration of [HA] = 
Then the pH of the buffer will be
pH = 6.247 + log [ 0.304/0.404]

So, the pH of the buffer is 6.1236.
Answer:
1) Ca: [Ar]4s²
2) Pm: [Xe]6s²4f⁵
Explanation:
1) Ca:
Its atomic number is 20. So it has 20 protons and 20 electrons.
Since it is in the row (period) 4 the noble gas before it is Ar, and the electron configuration is that of Argon whose atomic number is 18.
So, you have two more electrons (20 - 18 = 2) to distribute.
Those two electrons go the the orbital 4s.
Finally, the electron configuration is [Ar] 4s².
2) Pm
The atomic number of Pm is 61, so it has 61 protons and 61 electrons.
Pm is in the row (period) 6. So, the noble gas before Pm is Xe.
The atomic number of Xe is 54.
Therefore, you have to distribute 61 - 54 = 7 electrons on the orbitals 6s and 4f.
The resultant distribution for Pm is: [Xe]6s² 4f⁵.
Answer:
Part A
The volume of the gaseous product is
Part B
The volume of the the engine’s gaseous exhaust is
Explanation:
Part A
From the question we are told that
The temperature is 
The pressure is 
The of 
The chemical equation for this combustion is

The number of moles of
that reacted is mathematically represented as

The molar mass of
is constant value which is
So 

The gaseous product in the reaction is
and water vapour
Now from the reaction
2 moles of
will react with 25 moles of
to give (16 + 18) moles of
and 
So
1 mole of
will react with 12.5 moles of
to give 17 moles of
and 
This implies that
0.8754 moles of
will react with (12.5 * 0.8754 ) moles of
to give (17 * 0.8754) of
and 
So the no of moles of gaseous product is


From the ideal gas law

making V the subject

Where R is the gas constant with a value 
Substituting values
Part B
From the reaction the number of moles of oxygen that reacted is


The volume is


No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

Substituting values
Balanced equation:
Mg + 2 HNO3 —> Mg(NO3)2 + H2
This is a metal + acid reaction giving salt and hydrogen (not water).
The answer is A, do you want me to explain it? It’s pretty simple, you just need to follow all the signs in brackets and match them in those in the answer