Answer:
-1,103.39KJ/mol
Explanation:
We use the subtract the standard enthalphies of formation of the reactants from that of the products. It must be taken into consideration that the enthalpy of formation of elements and their molecules alone are not taken into consideration. Hence, what we would be considering are the standard enthalpies of formation of H2S, H2O and SO2.
In places where we have more than one mole, we multiply by the number of moles as seen in the balanced chemical equations.
The standard enthalpies of the molecules above are as follows:
H2S = -20.63KJ/mol
H2O = -285.8KJ/mol
SO2 = -296.84KJ/mol
O2 = 0KJ/mol
ΔrH⦵ = [2ΔfH⦵(H2O) + 2 ΔfH⦵(SO2)] − [ΔfH⦵(H2S) + 3
ΔfH⦵(O2)]
ΔrH⦵ =[(2 × -285.8) + (2 × -296.84)]
-[ 3 × -20.63)]
= (-571.6 - 593.68 + 61.89) = -1,103.39KJ/mol
Answer:
a. pH = 2 b. pH = 3 c. pH = 1 d. Unanswerable
Explanation:
pH = -log[H+] OR pH = -log{H3O+]
and inversely
pOH = -log[OH-]
1. Determine what substance you are working with, (acid/base)
2. Determine whether or not that acid or base is strong or weak.
a. 1.0 x 10^-2M HCl
HCl is a strong acid, therefore it will dissociate completely into H+ and Cl- with all ions going to the H+, therefore, the concentration of HCl and concentration of H+ are going to be equal, meaning we simply take the negative logarithm of the concentration of HCl and that would equal pH
pH = -log[H+]
pH = -log(1.0x10^-2)
pH = 2
b. 1.0 x 10^-3M HNO3
HNO3 like part a, is a strong acid, therefore it would simply require you to take the negative logarithm of the concentration of the compound itself, to find its pH.
pH = -log[H+]
pH = -log(1.0 x 10^-3)
pH = 3
c. 1.0 x 10^-1M HI
Like the previous parts, HI is a strong acid
pH = -log[H+]
pH = -log(0.10)
pH = 1
d. HB isn't an element, nor is it a compound so that would be unanswerable.
Answer:
Oxygen comes 8th on the periodic table which is the atomic number.
Explanation:
but Oxygen 13 (Isotopes of Oxygen) is when oxygen has 8 protons and electrons, and 5 neutrons (8+5=13)
1: a, 2:a, 3:a, 4:c, 5:d, 6:d, 7:c;
