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
Answer:
The answer is "the electron configuration of the element".
Explanation:
Electronics are distributed in atomic and molecular orbit via electrons from an atom or a molecule.
It reflects a most frequent dependence on valence electrons in the outer.
Through analyzing the context of the regular periodic table, the individual atoms are helpful. That's also important to understand chemical connections, which hold electrons together. This similar approach helps to explain the specific characteristics of lasers or semiconductors for bulk materials.
This is done by reducing each number by a common factor for each formula.
C3H6O6- all can be divided by 3 -> CH3O3 (you don't have to put a 1 for C, only if your teacher requests it. It is generally understood if not written)
H2O2- all can be divided by 2 -> H1O1 (had to put the 1's bc of the unintended language)
C8H8S2- all can be divided by 2 -> C4H4S
P5O15- all can be divided by 5- PO3
*****it is important to note that <em>all </em>numbers in the molecular formula are divided by the same thing to reduce them. decimals are <em>never </em>used in empirical formulas, only whole numbers.*****
