Answer:
Value for K would be too small. Less AgCl would dissolve due to the common ion effect due to the presence of Cl- in the water.
Explanation:
Think of this through the lenses of a shifting problem. Cl- ions are a product in this situation and increasing its concentration would shift the reaction back to the solid AgCl. In this specific case, due to Cl- ions, AgCl would dissolve less to maintain equilibrium and as a result, the concentrations of Ag+ and Cl- ions would be lower than normal making a smaller K value.
Since the electron is de-exited from 1(st) exited state (i.e n=2) to ground state (i.e n=1) for first line of Lyman series. Since the atomic number of Hydrogen is 1. λ = 4/3⋅912 A. 1/R = 912 A
The answers are a.) 0.03 mol KOH requires 0.03 mol HCl, b.) 2 mol NH3 requires 2 mol HCl and c.) 0.1 mol Ca(OH)2 requires 0.2 mol HCl.
Solution:
We need to write the balanced equations for each reactions to find out the stoichiometry for each reactants.
a.) HCl (aq) + KOH (aq) → KCl (aq) + H2O(ℓ)
From the balanced equations, we can see that 1 HCl reacts with 1 KOH, therefore if 0.03 mol KOH is reacted then 0.03 mol HCl must also be present.
b.) HCl(aq) + NH3(aq) ) → NH4Cl(aq)
If 2 moles of NH3 are reacted then 2 moles of HCl must also be present since 1 HCl reacts with 1 NH3 from the balanced reaction.
c.) 2HCl(aq) + Ca(OH)2(s) → CaCl2(aq) + 2H2O(ℓ)
We can see that 2 HCl react with 1 Ca(OH)2, hence if 0.1 mol of Ca(OH)2 is reacted then 0.2 mol HCl must also be present.
