Arrange the following solutions in order of decreasing freezing point: 0.20 m baf2, 0.15 m c6h12, 0.10 m crbr3, 0.15 m ch3ch2ch2cooh, 0.35 m kbr. (enter just the formulas in your answer. use the appropriate < , =, or > symbol to separate substances in the list.)

0.35 m > 0.20 m > 0.10 m > 0.15 m > 0.15 m

Explanation:

For the given aqueous solutions:

dissociates to form and

For 0.20 m of , the total concentration in ions = 3*0.20 = 0.60 m

does not dissociate. Therefore, the total concentration = 0.15 m.

dissociates to form and

For 0.10 m of , the total concentration in ions = 4*0.10 = 0.40 m

dissociates to form and .

For 0.15 m of , the total concentrations of ions = 2*0.15 = 0.30 m

dissociates to for and .

Thus, for 0.35 m of KBr, the total concentration of ions = 2*0.35 m = 0.70 m

Therefore, arranging the aqueous solutions in order of decreasing freezing point:

0.35 m > 0.20 m > 0.10 m > 0.15 m > 0.15 m

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

explanation: it gains negative electrons, therefore becoming negatively charged. as charge cannot be created, nor destroyed, only transferred, this leaves the wool positive. now that they have opposite charges, the two items attract.