Answer:
E = 1.602v
Explanation:
Use the Nernst Equation => E(non-std) = E⁰(std) – (0.0592/n)logQc …
Zn⁰(s) => Zn⁺²(aq) + 2 eˉ
2Ag⁺(aq) + 2eˉ=> 2Ag⁰(s)
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Zn⁰(s) + 2Ag⁺(aq) => Zn⁺²(aq) + 2Ag(s)
Given E⁰ = 1.562v
Qc = [Zn⁺²(aq)]/[Ag⁺]² = (1 x 10ˉ³)/(0.150)² = 0.044
E = E⁰ -(0.0592/n)logQc = 1.562v – (0.0592/2)log(0.044) = 1.602v
Answer:
The SAE curriculum includes practical farming tasks conducted outside the scheduled classroom and laboratory period by students. SAEs offer a method for students in agricultural education to gain real-world work opportunities that they are most interested in in the field of agriculture. Supervised agricultural experience is an essential component of agricultural education, and all Agriculture, Food and Natural Resources (AFNR) courses are a necessary component.
Explanation: Hope it helps
Answer: c. greater than 7.00
Explanation: The equivalence point of a titration is when all the base is consumed by the acid. When a strong base and a strong acid react, the medium is neutralized because is produced water and salt (which won't suffer hydrolysis). How water's pH is 7, in this type of titration the pH of the equivalence point will be at pH=7. But on titration of a weak acid with a strong base, the reaction of the equivalence point produces water and the conjugate base of the acid. Because the acid is weak, their conjugate base will be strong and will suffer hydrolysis, producing hydroxyl ions, elevating the pH of the water and making it greater than 7.
The Beer-Lambert law states that A = E*c*l where A is absorbance, E is the molar absorbance coeffecient, c is concentration and l is path length. Therefore the absorbance is directly proportional to concentration, and by increasing the concentration by a factor of 3, absorbance will increase by a factor of 3 giving A = 1.584
