"Hypotonic" refers to a solution that has a lower concentration of solutes compared to another solution. In biology, a hypotonic solution has a lower concentration of solutes outside of a cell compared to inside, causing water to move into the cell and potentially cause it to swell and burst.
A 25 millimole (m mol) solution means there are 25 millimoles of solute in every liter of solution. It is a unit used to express the concentration of a solute in a solution.
The Molarity is equal to the number of moles divided by the liters. M=mol/L M=0.120 mol/5.0 L M=0.024 so the molarity is 0.024 M
1.5M * 50mL / 0.4M = 187.5 mL = 190 mL
150 mL of a 2.5 M lidocaine solution contain 87,9 g.
"Hypotonic" refers to a solution that has a lower concentration of solutes compared to another solution. In biology, a hypotonic solution has a lower concentration of solutes outside of a cell compared to inside, causing water to move into the cell and potentially cause it to swell and burst.
The m/v of a solution is the mass of the solute divided by the volume of the solution, multiplied by 100.
No, the 0.75 M solution will have a higher freezing point. The freezing point depression is directly proportional to the molality of the solution, so a higher concentration solution will have a greater effect on lowering the freezing point.
A solution of NaCl 1 M.
The formula for calculating the molality (m) of a solution is: molality (m) moles of solute / kilograms of solvent.
It's simple to dilute a solution, just add more water. If you have one liter of 6 molar solution, and add another 5 liters of water, it becomes a 1 molar solution. You can also use M1V1 = M2V2 formula for dilution. For example, if you want to prepare 1 M of solution in 1 litre of water, then how much volume you need from the initial 6 M solution? Simply use the equation, (6 M) x (V1) = (1 M) x (1 litre) ---> V1 = (1 M) x (1 litre)/(6 M) = 0.167 L or 167 ml
To prepare a 0.50 M acetic acid solution, you would need to dilute the 2.5 M stock solution. By using the formula M1V1 = M2V2, you can calculate the volume of the stock solution needed as: (0.5 M)(100.0 mL) = (2.5 M)(V2), where V2 is the volume of the stock solution needed. So, V2 = (0.5 M x 100.0 mL) / 2.5 M = 20.0 mL. Therefore, you would need 20.0 mL of the 2.5 M stock solution to prepare the desired 100.0 mL of 0.50 M acetic acid solution.
The answer is 5 m L sodium bicarbonate, 1 M solution.
To find out how many milliliters of the 0.266 M LiNO3 solution are needed, you can use the formula C1V1 = C2V2, where C1 is the concentration of the first solution, V1 is the volume of the first solution, C2 is the concentration of the second solution, and V2 is the volume of the second solution. Plugging in the values, you can solve for V1, which will give you the volume of the 0.266 M LiNO3 solution needed to make 150.0 ml of 0.075 M LiNO3 solution.
A 25 millimole (m mol) solution means there are 25 millimoles of solute in every liter of solution. It is a unit used to express the concentration of a solute in a solution.
It would contain the same total ion concentration as 0.0750 M Na3PO4.
The pH of a solution containing 0.1 M of HC2H3O2 is around 2.88.