Why is it important to keep the temperature low when adding ethanol? Teacher said something about ethanol escaping?
Also, what factors affect size of crystal growth?
The temperature should be kept low as chromium (III) sulphate complexes form above about 60 °C. Some of these complexes are stable and the equilibria not easily reversible. For example, a chromium (III) sulphate complex is known which does not give characteristic the reactions of either Cr³+ or SO4²- in solution (e.g. sulphate with barium chloride).
Crystallisation is a process finely balanced bewteen thermodynamics and kinetics. Crystals form when the concentration of a solution is higher than the solubility product under stable conditions to maintain local supersaturation of the solution. However, to start crystallisation (which is an exothermic process) you need nucleation which is normally an endothermic process. Vibrations etc put energy into the system encouraging nucleation and once nucleation is complete crystal growth is accelerated (i.e. kinetics take over).
Crystallisation is affected by many other factors including relative and absolute concentrations of all components in the solution, method of crystallization and geometry of apparatus, pH of the solution etc.
Chemistry!! Please help! What mass of precipitate forms when 150.5 mL of 0.706 M NaOH is added to 517 mL...?
Ok, I have the equations I'm just struggling with the calculations. Please help!
Aluminum sulfate, known as cake alum, has a remarkably wide range of uses, from dyeing leather and cloth to purifying sewage. In aqueous solution, it reacts with base to form a white precipitate.
(a) Write balanced total and net ionic equations for its reaction with aqueous NaOH. (Type your answer using the format (NH4)2CO3 for (NH4)2CO3, [NH4]+ for NH4+, and [Ni(CN)4]2- for Ni(CN)42-. Use the lowest possible coefficients. Type the cation before the anion.)
total equation
1 Al2(SO4)3 (aq) + 6 NaOH(aq) >> 3 Na2SO4 + (aq) + 2 Al(OH)3 (s)
net ionic
1 [Al}3+ (aq) + 3 [OH]- (aq) >> 1 Al(OH)3 (s)
(b) What mass of precipitate forms when 150.5 mL of 0.706 M NaOH is added to 517 mL of a solution that contains 23.6 g aluminum sulfate per liter? _____________ g
2.76 g
****************
start here...remember these steps for problems like this
1) write balanced equation
2) convert everything to moles
3) determine limiting reagent
4) use balanced equation to convert from moles limiting reagent to moles of other species
5) convert moles back to mass
if we apply those steps to your problem
**** 1****
1 [Al]3+ (aq) + 3 [OH]- (aq) ---> 1 Al(OH)3 (s) is correctly balanced
**** 2 ***
M = molarity = moles / L right?
moles NaOH = 0.1505 L x (0.706 moles / L) = 0.1063 moles
mass Al2(SO4)3 = 0.517 L x (23.6 g Al2(SO4)3 / L) = 12.2012 g
molar mass Al2(SO4)3 = 342.15 g/mole
moles Al2(SO4)3 = 12.2012 g x (1 mole / 342.15 g ) = 0.03566 moles
since we're working with net ionic, let's convert moles NaOH and moles Al2(SO4)3 to moles OH- and moles Al+3... this isn't necessary though. we could just use the total equation...but that seems to be the direction this problem is heading so I'll go along with it ok?
