Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.
1) cobalt u can tell bc on a periodic table there is a small number that on cobalt is 27.
2) think that would be 11 bc in the 4th shell there can be up to 18 electrons
Noble gases:) they are very non-reactive
The fomula is NH4 (1+)
There are only two elements N and H.
As per oxidation state rules, the most electronegative element will have a negative oxidation state and the other element will have a positive oxidation state.
N is more electronative than H, so H will have a positive oxidation state and nitrogen will have a negative oxidation state.
You can also use the rule that states the hydrogen mostly has 1+ oxidation state,except when it is bonded to metals.
In conclusion the oxidation state of H in NH4 (1+) is 1+.
Now you must know that the sum of the oxidations states equals the charge of the ion, which in this case is 1+.
That implies that 4* (1+) + x = 1+
=> x = (1+) - 4(+) = 3-
Answer: the oxidation state of N is 3-, that is the option b.
Answer:
2Li(s) + ⅛S₈(s, rhombic) + 2O₂(g) → Li₂SO₄(s)
Explanation:
A thermochemical equation must show the formation of 1 mol of a substance from its elements in their most stable state,.
The only equation that meets those conditions is the last one.
A and B are wrong , because they show Li₂SO₄ as a reactant, not a product.
C is wrong because Li⁺ and SO₄²⁻ are not elements.
D is wrong because it shows the formation of 8 mol of Li₂SO₄.
