Let v be the capacity in milliliters of the 75% solution required then (90 - v) is the required capacity of the 84% solution needed.
75/100v + 84/100(90 - v) = 77/100 x 90
75v + 84(90 - v) = 77 x 90
75v + 7560 - 84v = 6930
9v = 7560 - 6930 = 630
v = 70 : therefore (90 - v) = 20
The mixed solution requires 70ml of the 75% solution and 20ml of the 84% solution to create 90ml of 77% solution.
Thales was not a chemist; important contributions in geometry and astronomy.
's called a MOLE
2.674 grams
If you were ill, and had to find out whether your temperature was raised or not, you would have to use decimals to judge this. (At least, if you were using Celsius you would need to use decimals). Likewise, if you were to use cookery thermometers to assess how "cooked" a steak or sugar syrups are. Alternatively, when using an electronic weighing scale to weigh out substances for which accuracy is crucial (e.g yeast, agar, emulsifiers, a chemist weighing drugs etc...) - all substances which must be weighed on a scale which can measure "0.001" of a gram. Capacity of car engines are measured in decimals - e.g 1.2 litre engine. If you have to calculate how much carpet you need to cover a room - the length, width or area are very unlikely to come out as a whole number.
33ml
nvm figured the answer out it's 70 ml of 75% ; 20 ml of 84%
x+y=110 .53x+.97y=.65*110 x=110-y .53(110-y)+.97y=.65*110 .53*110-.53y+.97y=.65*110 .97y-.53y=(.65-.53)*110 .44y=.12*110 y=.12*110/.44 = 30 ml x=110-30 = 80 ml So, you need 80 ml of the 53% solution and 30 ml of the 97% solution.
some liquid volumes are not additive, leading to potentially confusing final solution volumes.
For a detailed answer visit: algebra.com/algebra/homework/word/mixtures/Mixture_Word_Problems.faq.question.786696.html
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Become a chemist, or hire one.
You will need more data about all densities (in kg/Litre)and you must be sure of using mass% = (g solute)/(100 g solution)Solve two equations for both X and Y:4*d50*50 = X*d20*20 + Y*d70*70 (based on salt mass balance in diff. sol'n.)4*d50 = X*d20 + Y*d70 (based on solutions mass balance)In which:dm = density of the 'm'% salt solution in kg/Litre)X and Y = volume of the 20% and 70% salt solutions respectively
30 liters of a 10 % solution of fertilizer has .1(30) = 3 liters of fertilizer 1 liter of 30% solution has .3 liter of fertilizer 10 liters of 30% solution has 3 liters of fertilizer so, the chemist needs 10 liters of the 30% solution and 20 liters of water to make 30 liters of a 10% solution.
little solvent and much solute r.s.c
Stanley Rottiser Benedict, an American chemist
0.5 ml or 1/2