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9.46 grams
This depends on the volume and concentration of this solution.
I hope you don't work in a hospital, and I also hope you have passed 7th grade.But, 5% of a 1 L (1000 g) solution would be 50 grams.
26.8125 g
.05 L x 342 (this is the molecular weight of sucrose) x 1.75 = 29.925g
7.5 grams are needed because each milliliter is 1.50 grams, to get 5.00 mls, the calculation would be 5x1.50=7.5.
The needed mass is 35,549 g.
124,9 g grams of ammonium carbonate are needed.
290 grams
29.8g H2O = 1.66 mol H2O Molar Mass CuSO4 * 5H2O = 249.6 g mol CuSO4 * 5H2O --> 5 mol H2O 249.6 g CuSO4 * 5H2O/1 mol CuSO4 * 5H2O Times * 1mol CuSO4 * 5H2O/5mol H2O Times* 1.66 mol H2O = 82.6 g CuSO4 * 5H2O
29.8g H2O = 1.66 mol H2O Molar Mass CuSO4 * 5H2O = 249.6 g mol CuSO4 * 5H2O --> 5 mol H2O 249.6 g CuSO4 * 5H2O/1 mol CuSO4 * 5H2O Times * 1mol CuSO4 * 5H2O/5mol H2O Times* 1.66 mol H2O = 82.6 g CuSO4 * 5H2O
The sodium chloride mass needed is 292,2 g
400 mls would require 40g of glucose for a 10% solution and thus 20g for a 5% solution.
More than 45,5 g KNO3.
9.46 grams
Grams
Take 5 grams of calcium chloride and dissolve it in 100ml of solution to get a 5% solution of calcium chloride. The standard way to make a weight-volume solution is to take grams of the dry substance in 100ml of volume.