Answer:
The required volume of the original sample required is 2 micro liter
Explanation:
assuming the original sample concentration is 1N
after final dilution of 10-2 solution concentration becomes 0.01 N
normality of original sample = 1 N
normality of final solution = 0.01 N
volume of original sample= ?
volume of final solution = 0.2 mL
Considering thef formula below :
N1V1 = N2V2
V1 = (N2V2)/N1
= (0.01*0.2)/1
= 0.002 mL
1 milli liter = 1000 micro liter
0.002 mL = 2 micro liter
The original sample required is 2 micro liter
Answer:
Higher oxidation state metals form stronger bong with ligands
Explanation:
Ligand strength are based on oxidation number, group and its properties
I'm not sure, I think he'll mess everything because he can't see. Let me know if this helped ♂️
Answer:
ΔE(work)= 692 Kj/mole e⁻ = 4.32 x 10²⁴ eV/mole e⁻
Explanation:
ΔE = hc/λ = (6.63 x 10⁻³⁴j·s)(2.99792 x 10⁸m/s)/(1.73 x 10⁻⁷m) = 1.15 x 10⁻¹⁸j/e⁻ x 6.023 x 10²³ e⁻/mole = 691,999 j/mole e⁻ = 692 Kj/mole e⁻ x 6.24 x 10²¹ eV/KJ = 4.32 x 10²⁴ eV/mole e⁻
Answer:
The correct answer is The equilibrium constant for this reaction changes as the pH changes.
Explanation:
The equilibrium constant of a reaction depends of pH and viceverca.There is a equation that link pH with equilibrium constant Ka(for assumption).The equation is given below,
pH=pKa+log[A-/HA]
where[A-] is the concentration of conjugate base and [HA] is the concentration of conjugate acid.
