The ideal gas under STP is 22.4 L/mol. While the gas has a rule of P1V1/T1=P2V2/T2. So the volume under 101 kPa and 273 K is 0.2*22.4=4.48 L.
Answer:
the pH of HCOOH solution is 2.33
Explanation:
The ionization equation for the given acid is written as:

Let's say the initial concentration of the acid is c and the change in concentration x.
Then, equilibrium concentration of acid = (c-x)
and the equilibrium concentration for each of the product would be x
Equilibrium expression for the above equation would be:
![\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}](https://tex.z-dn.net/?f=%5CKa%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BHCOO%5E-%5D%7D%7B%5BHCOOH%5D%7D)

From given info, equilibrium concentration of the acid is 0.12
So, (c-x) = 0.12
hence,

Let's solve this for x. Multiply both sides by 0.12

taking square root to both sides:

Now, we have got the concentration of ![[H^+] .](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20.)
![[H^+] = 0.00465 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20%3D%200.00465%20M)
We know that, ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
pH = -log(0.00465)
pH = 2.33
Hence, the pH of HCOOH solution is 2.33.
Answer:
Electrons fill orbitals in order of increasing orbital energy
Explanation:
Apex