5 * 10**-12 mol
32 * 10**-9 mol
Concentration (M) * Volume (L) = mols
C1*V1=C2*V2
(5*10**-12)*V1=(32*10**-9)*V2
(5*10**-12)*V1/(32*10**-9)=V2
(5*10**-3)*V1/32=V2
The volume of the 5 picomolar solution that you wish take = V1
The volume of the 32 nanomolar solution that you need to make V1 at 5pM concentration = V2
Take V2, and place into graduated cylinder and fill to V1.
To prepare 100 ml of a 5% dextrose solution from a 50% dextrose solution, you would use the formula: C1V1 = C2V2. You will need 10 ml of the 50% solution (C1) and dilute it with 90 ml of water (V1) to obtain the desired 100 ml of 5% dextrose solution.
To prepare 1 liter of a 5% alcohol-water solution, you would mix 50 mL of alcohol with 950 mL of water. This ratio would give you a total volume of 1 liter with 5% of it being alcohol.
Denpending of the concentrations, you can use only mechanical agitation and elevate temperature for 5 to 15 minutes.
You can prepare a 2 mg/ml protein solution by diluting the 10 mg/ml protein solution with a diluent in a 1:5 ratio. Measure 8 ml of the 10 mg/ml protein solution and add 32 ml of the diluent to make a total volume of 40 ml. Mix properly to ensure uniform distribution of the protein in the solution.
To prepare a 5 M NaOH solution, measure out 200 g of NaOH pellets and dissolve them in enough water to make 1 liter of solution. To adjust the pH to 12, you can titrate the solution with a standard acid solution (e.g. HCl) until the desired pH is reached. Remember to wear appropriate safety gear and use a fume hood when working with NaOH.
To prepare a 5% potassium iodide solution, weigh 5 grams of potassium iodide and dissolve it in 100 mL of water. Stir until the potassium iodide is completely dissolved to achieve a 5% solution.
See the two Related Questions to the left for the answer.The first is how to prepare a solution starting with a solid substance (and dissolving it). The second question is how to prepare a solution by diluting another solution.
To prepare a 5% Dettol solution, you would mix 95 parts water with 5 parts Dettol. For example, to make 1 liter of 5% Dettol solution, you would mix 950 ml of water with 50 ml of Dettol. Always ensure to follow the manufacturer's instructions for proper dilution ratios.
To prepare 100 ml of a 5% dextrose solution from a 50% dextrose solution, you would use the formula: C1V1 = C2V2. You will need 10 ml of the 50% solution (C1) and dilute it with 90 ml of water (V1) to obtain the desired 100 ml of 5% dextrose solution.
To prepare a 5% NaCl solution, you will need 200 grams of NaCl for 4000 mL (4 L) of solution. This is calculated as 5% of 4000 mL, which equals 200 grams.
To prepare 1 liter of a 5% alcohol-water solution, you would mix 50 mL of alcohol with 950 mL of water. This ratio would give you a total volume of 1 liter with 5% of it being alcohol.
520 ml of HCl in 480 ml of water=1000ml = 5 N
400 mls would require 40g of glucose for a 10% solution and thus 20g for a 5% solution.
To prepare 1000 ml of 0.02 M NaCl solution, you would need 40 ml of 5 M NaCl solution, which you can calculate using the formula C1V1 = C2V2, where C1 is the concentration of the stock solution, V1 is the volume of the stock solution needed, C2 is the desired concentration, and V2 is the final volume. The dilution factor in this case would be 25, as you are diluting the 5 M solution 25 times to achieve the desired 0.02 M concentration.
Denpending of the concentrations, you can use only mechanical agitation and elevate temperature for 5 to 15 minutes.
125 ml 500(ml) * 0.05 = 25 25 / 0.20 = 125
10g/200ml=5g/100ml = 5% solution. Now read your Dosage and Calculations book and prepare for the test because it is Monday. Good luck!