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
your answer will be :
B. <u>Na has a lower</u> <u>electronegativity than H</u>
because Na belongs to alkali metals which are least electronegative (most electro positive) but hydrogen is a non metal, it has higher electronegativity as compared to metals like Sodium (Na).
Answer:
91.2 nm
Explanation:
The Rydberg equation is given by the formula
1/ λ = Rh ( 1/ n₁² - 1/ n₂²)
where
λ is the wavelength
Rh is Rydberg constant
and n₁ and n₂ are the energy levels of the transion.
We can see from this equation that the wavelength is inversely proportional to the difference of the squares of the inverse of the quantum numbers n₁ and n₂. It follows then that the smallest wavelength will be given when the the transitions are between the greatest separation between n₁ and n₂ whicg occurs when n1= 1 and n₂= ∞ , that is the greater the separation in energy levels the shorter the wavelength.
Substituting for n₁ and n₂ and solving for λ :
1/λ = 1.0974 x 10⁷ m⁻¹ x ( 1/1² -1/ ∞²) = 1.0974 x 10⁷ m⁻¹ x ( 1/1² - 0) =
λ = 1/1.0974 x 10⁷ m = 9.1 x 10⁻8 m = 91.2 nm
24518 is the answer to your queston
