300
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 answer is 1 471,5 mL.
164,7 mL are needed.
200 grams of solution will contain 200 x 4% or 200 x 0.04 = 8.0 total grams of solute.
The number of grams of ore that is needed to obtain 454.0 grams of Fe is 1297 grams. The answer was achieved by dividing 454.0 grams by 35 percent.
40.8 grams
It would be 12.6g of IKI to obtain the 100mL solution of 0.300 M IKI.
It would be 12.6g of IKI to obtain the 100mL solution of 0.300 M IKI.
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 answer is 0.8 L: Let's think of a '6/15 solution' as meaning 6/15 grams per litre, and the solution we're trying to obtain is 10 grams/litre. In our final solution, we'll have 6 grams resulting from the one litre of weaker solution, and 15v grams from the stronger (where v is the volume of stronger added). Deviding this total mass by the total volume of the solution, we arrive at the wanted concentration of 10 grams/litre. Algebraicly: (15v + 6)/(1+v) litres = 10 15v + 6 = 10 + 10v 5v = 4 v = 4/5 = 0.8 L Hope this helps.
The answer is 364 mL.
The answer is 1 471,5 mL.
164,7 mL are needed.
mixture
489 grams
The percent concentration is 13,75 %.
This is (mass of solute) divided by (mass of total solution) expressed as a percentage. The solute is what you are dissolving into the solution. Example: you have 90 grams of water, and you add 10 grams of salt (sodium chloride). The water is the solvent, sodium chloride is the solute, and the solution is salt water. 90 grams + 10 grams = 100 grams (mass of total solution). (10 grams) / (100 grams) = 0.1 --> 10% mass mass percent concentration.