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How do you prepare 1millimole of silver- nitrate solution?
To prepare a 1 millimole (1 mM) solution of silver nitrate, you would dissolve 0.168 grams of silver nitrate (AgNO3) in enough water to make 1 liter of solution. The molar mass of AgNO3 is 169.87 g/mol, so 1 millimole is 0.169 grams. Be sure to use proper lab techniques and safety precautions when handling chemicals.
How many grams needed from water to prepare 160 grams from potassium acetate 5 percent w?
To prepare 160 grams of potassium acetate with a 5% w/w concentration, you would need to calculate the mass of the potassium acetate in the solution. Since the concentration is given as a percentage by weight, 5% of 160 grams is 8 grams of potassium acetate. The remaining mass in the solution would be water. Therefore, you would need 152 grams of water to prepare 160 grams of potassium acetate with a 5% w/w concentration.
How much gram in a millimole?
The number of grams in a millimole depends on the molecular weight of the substance being measured. To calculate it, you need to know the specific molecular weight of the substance.
How many grams of dextrose are needed to prepare 0.5l of d5w?
To prepare 0.5L of D5W (5% dextrose in water), you need 25 grams of dextrose. This is because 5% of 0.5L is 25 grams.
How many grams in one millimole?
The number of grams in one millimole depends on the specific substance being considered. To find the mass of one millimole of a substance, you would need to know its molar mass, which is typically expressed in grams per mole. You can then simply divide the molar mass by 1000 to get the mass of one millimole in grams.
How many grams of Na OH are needed to prepare 75.0 g of Na OH?
75g
How much sodium chloride in grams is needed to make 13 L of a solution that has a concentration of 4 grams of sodium chloride per liter of solution?
To calculate the total amount of sodium chloride needed for a 13 L solution at 4 grams per liter, multiply the concentration by the volume of the solution: 4 grams/L x 13 L = 52 grams of sodium chloride. Therefore, you will need 52 grams of sodium chloride to make the 13 L solution.
How many grams of sodium metabisulphite is needed to make 0.01N solution?
To prepare a 0.01N solution of sodium metabisulfite, you would need 2.31 grams of sodium metabisulfite per liter of solution.
How many grams of NaOH are needed to prepare 250mL of 205 M of NaOH?
To calculate the grams of NaOH needed, you first need to find the moles of NaOH required: 205 M concentration means 205 moles/L. So, for 250 mL (0.25 L), you multiply 0.25 L by 205 moles/L to get 51.25 moles. Finally, using the molar mass of NaOH (40 g/mol), you can convert moles to grams by multiplying 51.25 moles by 40 g/mol to find you need 2050 grams of NaOH.
How do you prepare 1 liter of 100ppm NaCl solution?
To prepare a 1 liter of 100ppm NaCl solution, you would dissolve 0.1 grams of NaCl in 1 liter of water. This concentration is achieved by mixing 0.1 grams of NaCl in 1 liter of water.
How many grams of KCl are needed to prepare 125 mL of a 0.720 M KCl solution?
To calculate the amount of KCl needed, we first need to find the number of moles of KCl required using the formula: moles = Molarity x Volume (in L). Then, we convert moles to grams using the molar mass of KCl, which is 74.55 g/mol. Finally, we use the formula: grams = moles x molar mass to find that approximately 6.33 grams of KCl are needed to prepare 125 mL of a 0.720 M solution.
What is the mass percent concentration for 5 grams of salt in 75 grams of water?
5 grams of salt in 75 grams of water = 5 grams of salt in 80 grams of the solution.So the mass concentration = 5/80 = 100*5/80% = 6.25%5 grams of salt in 75 grams of water = 5 grams of salt in 80 grams of the solution.So the mass concentration = 5/80 = 100*5/80% = 6.25%5 grams of salt in 75 grams of water = 5 grams of salt in 80 grams of the solution.So the mass concentration = 5/80 = 100*5/80% = 6.25%5 grams of salt in 75 grams of water = 5 grams of salt in 80 grams of the solution.So the mass concentration = 5/80 = 100*5/80% = 6.25%
