i wonder if you could freeze part A then put part B over it let it freeze.. and dose both parts at once....
i wonder if you could freeze part A then put part B over it let it freeze.. and dose both parts at once....
Well salt lowers the freezing point of water and Randy's recipe has 86,000 ppm chloride in the mag supplement. So it's very salty. I'm sure the calcium supplement is also very salty. I'm not sure at what temp they freeze, but I'm not sure you could get your freezer that cold.
Calcium chloride CaCl2 can melt ice at -29°C (-20°F)
Magnesium chloride MgCl2 at -15°C (5°F)
Now Randy's recipe takes the CaCl2 and MgCl2 and dissolves it in water which 'dilutes' (not really the right word...reduces it's potency perhaps) it somewhat I'm sure. Maybe it would be possible to freeze the magnesium chloride.
i dont see the benefit of freezing the calcium since you can dose lots of calcium at once with no ill effect
the alkalinity supplement maybe.. especially if you need to cool your tank off with ice anyways.
The calcium chloride freezes as well or did you just try the alk supplement?
i dont see the benefit of freezing the calcium since you can dose lots of calcium at once with no ill effect
the alkalinity supplement maybe.. especially if you need to cool your tank off with ice anyways.
I agree.
Besides I've taken up enough room in the freezer with food and stuff.
And not to mention avoiding that embarassing,"these ice cubes taste funny"situation.
if i did my calculations correctly, which i might not of... but the freezing point of randy's 2 part calcium chloride should be -6.6 degrees C, which is equivalent to 20 degrees Fahrenheit... a temperature that freezers can reach.
So my thinking is yes you could freeze it no problem, but i still don't think the whole idea would work that well... you'd still need to measure out volumetric amounts of the 2 part solutions into the ice cube trays to know how much you're dosing with each ice cube... and that's only if the whole dosing as a solid thing would work.. which i'm not sure if it would.
formula used was delta T= Kf (of water, the solvent) x N (number of ions in solution) x m (molality of the solution)
molality was calculated using moles CaCl2/Kg of water, so 4.5/3.8= 1.1842
so delta T = 1.86 x 3 (CaCl2) x 1.1842 (the molality of the solution)
Delta T= 6.6 degrees C