Calculate the milligrams of metallic ion that can remain at equilibrium in a solution of Fe(OH)3 having a [OH-] = 1.0 x 10-4 mol/L and Ksp = 6.0 x 10-38.
yes that is what i meant, my apologies
by combine do you just mean say that there is 5.98 kJ per 0.0260 moles of MgO
or 5.98kJ/0.0260 mol MgO= 230kJ/mol
ok, thanks, i see so it must be 5.98KJ, makes sense
for the second part, hmmm how about if i did the amount of MgO which is given
(1.05g MgO)(1mol MgO)/(40.31 g MgO)=.0260
im not sure about this one
thanks, so how about i transfer the 110.15ml to grams which is 110.15g and then do q=mc\deltat
q=(110.15g)(4.18J/g^oC)(13^oC)
q=5985.551 J = 598 KJ
q=5.9x10^3 J
so the answer is q= 598 KJ or q=5.985x10^3 J ??
better?
also how can I calculate KJ/mol of MgO from this? has to do with the molar...
1.05 g MgO is added to 110.15ml HCL with an initial temperature of 21.5 degrees Celsius and a final temperature of 34.5 degrees Celsius. calculate the KJ of energy released for the reaction. assume that 4.18 J of energy is required to change the temperature of one mL of solution to one Celsius...