694 mL of glucose solution 0,2 M are needed.
moles glucose needed = 25.0 g x 1mol/186 g = 0.139 moles
voluem of 0.2 M needed: (x L)(0.2 mol/L) = 0.139 moles
x = 0.694 L = 694 mls
30 ml 50% dextrose + 70 ml water
25
25
You are missing a measurement to obtain volume. A height it going to be needed.
BY adding excess Filter off excess MgCO3(s) . heat the filter to get a saturated solution , cool to obtain crystals , filter . H2SO4(aq)+MgCO3(s)--->MgSO4(aq)+H2CO3(g)
pure
When you use your senses to obtain information, you are making an observation. This is a major part of science where you use your senses or the readings of instruments to directly acquire information from a primary source.
Temperature and gravity.
A pharmacist mixed a 20 percent solution with a 30 percent solution to obtain 100 liters of a 24 percent solution. How much of the 20 percent solution did the pharmacist use in the mixture (in liters).
10
600ml.
25 gallons
10 liters.
If the percents given are by weight or mass, this is very straightforward: The ratio between the desired percentage and the initial percentage is 1/50. Therefore, a given mass of initial solution must be diluted to 50 times its original mass to obtain the desired lower concentration, or in other words, 49 parts of diluent must be mixed with each part of initial solution. If the percents involve volume measurements, it would be necessary to take into account and change in density occasioned by the dilution.
4 litres
50 gallons @ 3% must be added.
40.8 grams
0.6 of a pint.
80% water
28 milliliters