Answer: 1.2642*10²⁵ on both sides
Explanation:
First check how many moles are there on each side.
Since this is a balanaced equataion the number of moles on each side is the same thus the number of atoms is also same on both sides
There are 3 moles of carbon and 8 moles of hydrogen in C3H8
and 2 moles of oxygen in O2 but there 5 infront so 2*5 is 10
Number of moles on the right is 10+8+3 = 21
Now use Avogrado's constant
21 Moles* (6.02*10²³)/Mol
= 21*6.02*10²³
= 1.2642*10²⁵
It is energetically favorable for all atoms to have a complete outer
electron shell. Loosely, the atoms on the left hand side of the periodic
table only have a few extra electrons in their outer shell so it is
energetically favorable for them to lose them. The atoms on the right
hand side of the periodic table almost have enough electrons in their
outer shell and so they have a tendency to gain them.
Once electrons have left an electron shell, an atom will have a positive
charge because it has more protons (positive charges) than electrons
(negative charges). Similarly, an electron which has gained electrons to
complete its outer shell will have a negative charge because it now has
more electrons (negative charge) than protons (positive charge).
Always remember that pH + pOH = 14
Here, you have a pOH of 11.24, so you replace it in the equation, and u get:
pH + 11.24 = 14
Then, You move 11.24 to the other part. and moving from a part to another change the sign of the equation. And you get:
pH = 14 - 11.24 = 2.76
So, the pH of a solution that has a pOH of 11.24 is pH = 2.76
Hope this Helps :)
Answer:
ΔH°rxn = - 433.1 KJ/mol
Explanation:
- CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)
∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state
∴ ΔH°CCl4(g) = - 138.7 KJ/mol
∴ ΔH°HCl(g) = - 92.3 KJ/mol
∴ ΔH°CH4(g) = - 74.8 KJ/mol
⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)
⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol
⇒ ΔH°rxn = - 433.1 KJ/mol
