Consider the formation of the three solutions shown in the table.

Type of Interaction Strength of Interaction
Solution A Solution B Solution C
solute-solute weak weak strong
solvent-solvent weak strong strong
solute-solvent strong strong weak

Rank the formation of the solutions A, B, and C from the most exothermic to the most endothermic. Rank the enthalpy of solution from most exothermic to most endothermic. To rank items as equivalent, overlap them.

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

d.) circle motion requires a force to give it a centripetal acceleration.

i wish i could but this one is weird give me abc and i can solve it.