1N HCL is the same as 1 Molar HCl.
You take the # of H ions and multiply by the molarity to get the Normality.
Usually you buy HCl in concentrated form which is 12 Molar or 12 Normal HCL.
You need to dilute the concentrated HCl to get the reduced concentration.
Use the formula Molarity Initial x Volume Initial = Molarity Final x Volume Final
ex. 12 M HCL x 10 ml = 1 M x 120 ml.
So take 10 ml of concentrated HCl and add enough water to make 120 ml. This will give you 120 ml of 1 M (which is 1N) HCl.
Venkat Reddy
To prepare a molar solution of iron(III) sulfate, first calculate the molar mass of Fe2(SO4)3. Then, dissolve the calculated mass in a known volume of solvent, typically water, to achieve the desired molarity. Finally, ensure complete dissolution and accurate volume measurement for a precise molar solution.
To find the amount of chlorine used, we need to know the molar mass of chlorine. We can use the molality to calculate the moles of chlorine in the solution. Finally, we can convert moles to grams using the molar mass of chlorine to find the grams of chlorine used.
To prepare 3% hydrogen peroxide from 30% hydrogen peroxide, you need to dilute the 30% solution with water. The formula to calculate the dilution is C1V1 = C2V2, where C1 is the initial concentration (30%), V1 is the volume of the initial solution, C2 is the final concentration (3%), and V2 is the final volume of the solution you want to prepare. By plugging in the values, you can determine the volume of 30% hydrogen peroxide needed and the volume of water needed to achieve a 3% solution.
To make a 1M solution of sodium chloride in 1 liter of water, you would need 58.44 grams of sodium chloride. This is based on the molecular weight of sodium chloride, which is 58.44 g/mol.
To make a 1 molar solution of sodium azide, you would need to dissolve 65.01 g of sodium azide in water to make 1 liter of solution. Since you have 98 mg of sodium azide, you would need to add enough water to make a final volume of 1 liter to create the 1 molar solution.
To prepare a 500mM KCl solution, you would need to dissolve 74.55 grams of KCl in enough solvent to make 1 liter of solution.
To prepare a 1 normal solution of NaOH, you would need to dissolve 40.00 grams of NaOH in enough water to make 1 liter of solution. This is because the molecular weight of NaOH is 40 g/mol.
2 lt
To prepare a 40.0% lithium nitrate solution, 60mL of the solution would be solvent (water) and 90mL would be solute (lithium nitrate). Therefore, you would need 90mL of lithium nitrate to prepare 150mL of the 40.0% solution.
Image result for You prepare a less concentrated H C l solution from a stock solution with 12m concentration. If you too 100g of the stock solution to prepare 4 MHCl solution how much water is needed to prepare o find solution 9density HCL(12) = 1,89/ml? The concentration would be 0.76 mol/L.
it required too much time to prepare new craft
To prepare a 0.05 M disodium EDTA solution, you would need to dissolve 3.72 grams of disodium EDTA dihydrate (Na2C10H14N2Na2·2H2O) in enough water to make 1 liter of solution.
To prepare a 500 ppm (parts per million) solution of lead nitrate, you would need to dissolve 500 grams of lead nitrate in 1 million grams (1,000,000 grams) of solution. The specific weight of lead nitrate would depend on its molar mass and density. It is more common to express concentrations in terms of molarity (moles per liter) rather than ppm for making solutions.
Approximately 18.4 grams of baking soda is required to make 50 ml of a saturated solution at room temperature.
RMM of CuSo4 . 5H2O = 160 + 5x18 = 250 g How much contain gram in 100 mL of 0.050 M CuSO4 solution is - Solution: = 0.050 x 100/1000 x 250 = 0.050x4 (100)/ 16 = 1.25 g
You would need to use 17.5 mL of the 0.250 M sucrose solution to prepare 400.0 mL of a 0.0310 M solution. This can be calculated using the formula: C1V1 = C2V2, where C1 is the initial concentration, V1 is the volume of the initial solution, C2 is the final concentration, and V2 is the final volume.
To prepare a 1000 ppm (parts per million) solution of KMnO4 (potassium permanganate), you need 1000 mg of KMnO4 per liter of solution. Since 1 gram equals 1000 mg, you would need 1 gram of KMnO4 dissolved in enough water to make a final volume of 1 liter. Therefore, to prepare a 1000 ppm solution, dissolve 1 gram of KMnO4 in 1 liter of water.