The concentration and volume of reagents combined in each trial are listed in the table below:
Trial 00020 M 00020 M H2O(mL)
Fe(NO3)3(mL) KSCN (mL)
1 5 2 3
2 5 3 2
3 5 4 1
4 5 5 0
Initially (right after combining and before any reaction occurs) the concentration of Fe3+ in trial 2 is
____M.
k so so far i have this can someone tell me if this is right...
3 ml Fe(NO)3 x (.0002 moles Fe(NO)3/ 1000ml Fe (NO)3) x (1 mole Fe (III)/ 1 mole Fe (NO)3)= .0000006 mol Fe (III)
then i think we are supposed to divide by total volume but i dnt know if it would be total volume for just trial 2 or the whole chart?
nevermind i got it
whats the anwser?
Your calculation of the moles of Fe3+ in trial 2 is correct. Now, let's move on to calculating the concentration of Fe3+ in trial 2.
To determine the concentration, you need to divide the moles of Fe3+ by the volume of the solution. In this case, since we are only interested in trial 2, we will divide by the volume of trial 2.
The volume of trial 2 is the sum of the volumes of Fe(NO3)3 and KSCN, which is 5 mL + 3 mL = 8 mL.
Now, let's calculate the concentration of Fe3+ in trial 2:
Concentration = moles of Fe3+ / volume of trial 2
Concentration = 0.0000006 mol / 0.008 L (since 8 mL is equal to 0.008 L)
Concentration = 0.075 M
So, the concentration of Fe3+ in trial 2, right after combining and before any reaction occurs, is 0.075 M.