Year 12s: what will be the first way you research your future options? >>

calculate initial conc

is the answer part of part b (i) 0.05 because of dividing the initial concentration of either, (which are the same,0.1) by 2? If so, if initial reactant concs of reactants are the same, do we ALWAYS divide by 2 for ANY reaction? or are there specific reactions when we should do this. image-9dcb1da9-69a4-4a34-ac8d-cb9aedc20110-931313105-compressed.jpg.jpeg

image-9dcb1da9-69a4-4a34-ac8d-cb9aedc20110-931313105-compressed.jpg.jpeg

Reply 1

Pigster

You are doubling the volume, so conc will halve.

Reply 2

sarah99630

Original post by Pigster

You are doubling the volume, so conc will halve.

thanks a lot! Oh my, u see in my notes it says conc doubles, volume halves. Now is when I realise its actually coz of the formula moles = C × V/1000
now it's easier to process cause I understand WHY.

Reply 3

sarah99630

Original post by Pigster

You are doubling the volume, so conc will halve.

Please could you help me understand something on the next part, which says calculate the number of moles of silver ions in the 10cm^3 sample at equilibrium. And the answer was found by multiplying 5.60 × 0.0200 and diving the answer by 1000, so I was wondering isn't this the moles that REACTED? cause all the calculations I've done before involve initial moles minus moles that reacted to get moles at equilibrium. (sorry for the chain questions)thanks again.

Reply 4

Pigster

Initially, you put Ag+ in. Some of that gets converted: Ag+ + Fe2+ <-> Ag + Fe3+. Some Ag+ remains in the equilibrium.

The equilibrium Ag+ reacts with the SCN- ions, in the equation given.

Reply 5

sarah99630

Original post by Pigster

Initially, you put Ag+ in. Some of that gets converted: Ag+ + Fe2+ <-> Ag + Fe3+. Some Ag+ remains in the equilibrium.

The equilibrium Ag+ reacts with the SCN- ions, in the equation given.