A. 16 of 18 percent and 2 of 9 percent b. 14 of 18 percent and 4 of 9 percent c. 16 of 9 percent and 2 of 18 percent d. 14 of 9 percent and 4 of 18 percent
Because the environment is Isotonic (having the same or equal osmotic pressure).
Set up. .70(2 gal) + X = .90(X + 2 gal) distribute 1.4 + X = .90X + 1.8 X = .90X + .4 .1X = .4 X = 4
add 4 parts water per part solution
4 ounces
A 2% salt solution is hypotonic compared to a 4% salt solution because it has a lower concentration of salt. In osmosis, water flows from hypotonic to hypertonic solutions, so in this case, water would flow from the 2% solution to the 4% solution to try to equalize the concentrations.
A. 16 of 18 percent and 2 of 9 percent b. 14 of 18 percent and 4 of 9 percent c. 16 of 9 percent and 2 of 18 percent d. 14 of 9 percent and 4 of 18 percent
To achieve a 50% salt solution with 4 liters total volume, the chemist needs 2 liters of each of the 20% and 70% solutions. This is because the resulting solution will be a combination of the two strengths in equal amounts, leading to an overall concentration of 50%.
Beggars and Choosers - 1999 Fifty Three Percent Solution 2-2 was released on: USA: 4 July 2000
Because the environment is Isotonic (having the same or equal osmotic pressure).
To make a two mole salt solution in two liters of water, you'll need 4 moles of salt (2 moles/L * 2 L = 4 moles). The molar mass of salt (NaCl) is approximately 58.44 g/mol, so 4 moles would be 233.76 grams (4 moles * 58.44 g/mol = 233.76 g).
1. Grinding of the rock salt. 2. Dissolving of the rock salt in water. 3. Filtering of the solution. 4. Repeated processes of crystallization/recrystallization.
Examples are:1- Cold drink 2- Air 3- Milk 4- blood 5- Sugar solution
4 percent
chemicals (45 percent), road de-icing/ice control salt (31 percent), salt sold to distributors (8 percent), industrial uses (8 percent), agricultural salt (6 percent), food (including table salt, 4 percent)
To make a 10 percent solution, you would need to dilute the 50 percent solution by adding 4 ml of solvent to 1 ml of the 50 percent solution. This will result in a total volume of 5 ml with a 10 percent concentration.
Set up. .70(2 gal) + X = .90(X + 2 gal) distribute 1.4 + X = .90X + 1.8 X = .90X + .4 .1X = .4 X = 4