The concentration of the OH- is 8.0 x 10-4 In terms of molarity, since the hydroxide is double that of the calcium, double the molarity of the solution.
1 mole of mass of ca (OH) 2 = 40*1 + (16 + 1) 2 = 40 + 34 = 74g. So, mass of 24 moles = 24*74 = 1776g.
The OH concentration in a 4.0 x 10^4 M solution of Ca(OH)2 can be determined by dividing the concentration of Ca(OH)2 by its stoichiometric coefficient, which is 2. Thus, the OH concentration is 2.0 x 10^4 M.
Molarity = moles of solute/Liters of solution ( 300.0 ml = 0.300 liters ) ( Ca(OH)2 is correct formulation ) 0.115 M Ca(OH)2 = moles Ca(OH)2/0.300 Liters = 0.0345 moles Ca(OH)2 (74.096 grams/1 mole Ca(OH)2) = 2.56 grams of Ca(OH)2 needed
Ca(OH)2 + Na2CO3 = CaCO3 + 2 NaOH
The atomic weight of Na-OH is 23+16+1= 40. So to prepare the one N solution of Na-OH, you need to add 40 grams of Na-OH in one litre of water. To prepare the 4N Na-OH solution, you need to add 160 grams of Na-OH in one litre of water.
1 mole of mass of ca (OH) 2 = 40*1 + (16 + 1) 2 = 40 + 34 = 74g. So, mass of 24 moles = 24*74 = 1776g.
The OH concentration in a 4.0 x 10^4 M solution of Ca(OH)2 can be determined by dividing the concentration of Ca(OH)2 by its stoichiometric coefficient, which is 2. Thus, the OH concentration is 2.0 x 10^4 M.
Molarity = moles of solute/Liters of solution ( 300.0 ml = 0.300 liters ) ( Ca(OH)2 is correct formulation ) 0.115 M Ca(OH)2 = moles Ca(OH)2/0.300 Liters = 0.0345 moles Ca(OH)2 (74.096 grams/1 mole Ca(OH)2) = 2.56 grams of Ca(OH)2 needed
Ca(OH)2 and Na2O
The mass of 7,346 moles of Ca(OH)2 is 544,3 g.
2 moles of Ca and 4 moles of OH
Ca(OH)2 + Na2CO3 = CaCO3 + 2 NaOH
The atomic weight of Na-OH is 23+16+1= 40. So to prepare the one N solution of Na-OH, you need to add 40 grams of Na-OH in one litre of water. To prepare the 4N Na-OH solution, you need to add 160 grams of Na-OH in one litre of water.
Ca(OH)2 is Calcium hydroxide (by chemical name), also known as lime, slaked lime, slack lime or pickling lime (by trivial and geological name)Limewater is the name of saturated solution of it in water.
m sub Ca(OH)2 = ( 10 g Ca ) [ ( 74.12 g Ca(OH)2 ) / ( 40.078 g Ca) ] m sub Ca(OH)2 = 18.5 g Ca(OH)2 <------------------
Ca(OH)2 and Mg(OH)2 react with sulfuric acid to form the salts, calcium sulfate and magnesium sulfate respectively
3,7 moles of Ca(OH)2 equals 274,1441 g. For 692 mL the concentration is 396,16 g/L.But this is only a theory because the solubility of Ca(OH)2 in water is very low and this solution cannot exist.