90 ml of dextrose and 4.41 litres of water.
125 ml 500(ml) * 0.05 = 25 25 / 0.20 = 125
400 mls would require 40g of glucose for a 10% solution and thus 20g for a 5% solution.
150 mL x 40.0 g LiNO3/100 mL solution = 60.g of solute
My calulation says 286ml of 25% and 214ml of 60% 286 x .25= 71.50 214 x .60= 128.40 250 x .20= 50.00 250 x .60= 150.00
8.0 g according to my teacher but i dont know the formula
Concentration of NaOH = 0.025 M = 0.025 Moles per Litre of SolutionVolume of Solution required = 5.00LWe can say therefore that:Number of Moles of NaOH needed to prepare the solution= Concentration of NaOH * Volume of Solution requiredTherefore:Number of Moles of NaOH needed to prepare the solution= 0.025M * 5.00L= 0.125molesFrom this we can say that 0.125 moles of NaOH are needed to prepare a 5.00 L solution with a concentration of 0.025M of NaOH.
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