Answer:
Sulfur: -1
Carbon: 0
Nitrogen: 0
Explanation:
The thiocyanate ion SCN- can have only two resonance structures, which are:
S - C ≡ N <--------> S = C = N
In the first structure, we have one single bond and one triple bond, in this case, the negative charge is located in the sulfur. This is because Sulfur have 6 electrons and those electrons are present in the atom, (see picture below), and counting the electron that is sharing with the Carbon, the total electrons that sulfur has is 7 (It has one more than usual). Carbon and nitrogen are already stable with 0 of formal charge, because carbon can only have 4 electrons which 1 is sharing with sulfur and the other 3 with the nitrogen, and nitrogen have 5 electrons, three sharing with carbon and the other two kept it for itself.
In the second structure, the negative charge of the sulfur is transfered to the nitrogen, meaning that it has 6 electrons the nitrogen (formal charge -1) and carbon and sulfur with 4 and 6 electrons respectively.
Between these two structures, the most stable is the first one basically because Sulfur is a better nucleophile than the Nitrogen, and can form stronger hydrogen bond in acid, giving more stable structure.
Answer:
I would put, A substance has two different parts of it, a pure substance and a compound and mixtures have two as well called, homogenous mixtures and hetrogeneous mixtures.
Explanation:
Answer:
Chemical compounds all have different melting points.
Explanation:
chemical compounds all have different freezing and boiling points. Different chemical compounds means they will have different chemical structures.
2 Na + 2 H2O → 2 NaOH + H2 (balanced equation)
The answer would be 2, since 2 in the coefficient of both Na and NaOH
Answer:
Final molarity of iodide ion C(I-) = 0.0143M
Explanation:
n = (m(FeI(2)))/(M(FeI(2))
Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol
So n = 0.981/309.85 = 0.0031 mol
V(solution) = 150mL = 0.15L
C(AgNO3) = 35mM = 0.035M = 0.035m/L
n(AgNO3) = C(AgNO3) x V(solution)
= 0.035 x 0.15 = 0.00525 mol
(AgNO3) + FeI(3) = AgI(3) + FeNO3
So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol
C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M