Since the volume ratio of two gases in a chemical reaction is directly proportional to the mole ratio of the reactants, you can infer that the mole ratio of lead nitrate to sodium iodide in their reaction is the same as the volume ratio of the gases involved. This allows you to determine the stoichiometry of the reaction.
By definition, a 0.61M sodium nitrate solution contains 0.61 moles of sodium nitrate per liter, which is equivalent to 0.61 mmol/ml. Therefore, the volume of this solution required to contain 400mmol is 400/0.61 or 6.6 X 102 ml, to the justified number of significant digits.
Thiosulfate can be standardised by adding excess potassium iodide solution to a known volume of a standard acidified solution of potassium dichromate, and then titrating the liberated iodine against the sodium thiosulfate solution.http://www.meduniv.lviv.ua/files/kafedry/tokshim/English/Analytical/Manuals_Analytchem/Iodometry.pdf
The volume of 24 mmol of ethyl iodide would depend on its density, which is about 2.29 g/cm³. To calculate the volume, you would need to convert 24 mmol to grams using the molar mass of ethyl iodide (155.99 g/mol) and then divide by the density. The theoretical yield of p-ethoxynitrobenzene would depend on the reaction conditions and stoichiometry of the reaction involving ethyl iodide, but you would use the molar ratio of ethyl iodide to p-ethoxynitrobenzene to calculate the theoretical yield.
To calculate the molarity, first convert the mass of sodium nitrate to moles using its molar mass. Then, divide the number of moles by the volume of solution in liters (265 mL = 0.265 L) to get the molarity. The molarity of the solution is about 0.68 M.
Dilute 1 mL of 0.5 M silver nitrate solution to a total volume of 1 L with water to make a 1 mM silver nitrate solution.
By definition, a 0.61M sodium nitrate solution contains 0.61 moles of sodium nitrate per liter, which is equivalent to 0.61 mmol/ml. Therefore, the volume of this solution required to contain 400mmol is 400/0.61 or 6.6 X 102 ml, to the justified number of significant digits.
To prepare a 1000 ppm nitrate solution from sodium nitrate, you would dissolve a specific amount of sodium nitrate in a known volume of water. For example, to prepare 1 liter of a 1000 ppm nitrate solution, you would dissolve 1 gram of sodium nitrate in 1 liter of water. Make sure to accurately weigh the sodium nitrate and dissolve it completely in the water to achieve the desired concentration.
The volume of 24 mol of ethyl iodide is 1,929 L.
Thiosulfate can be standardised by adding excess potassium iodide solution to a known volume of a standard acidified solution of potassium dichromate, and then titrating the liberated iodine against the sodium thiosulfate solution.http://www.meduniv.lviv.ua/files/kafedry/tokshim/English/Analytical/Manuals_Analytchem/Iodometry.pdf
To find the concentration of nitrate ions in the mixture, we first calculate the moles of nitrate ions from each salt using the formula moles = concentration x volume. For potassium nitrate: 0.120 mol/L x volume = moles of nitrate ions. For iron (II) nitrate: 0.160 mol/L x volume = moles of nitrate ions. Add the moles of nitrate ions from both salts to get the total moles of nitrate ions in the mixture. Finally, divide the total moles of nitrate ions by the total volume of the mixture to get the concentration of nitrate ions in the mixture.
The volume of 24 mmol of ethyl iodide would depend on its density, which is about 2.29 g/cm³. To calculate the volume, you would need to convert 24 mmol to grams using the molar mass of ethyl iodide (155.99 g/mol) and then divide by the density. The theoretical yield of p-ethoxynitrobenzene would depend on the reaction conditions and stoichiometry of the reaction involving ethyl iodide, but you would use the molar ratio of ethyl iodide to p-ethoxynitrobenzene to calculate the theoretical yield.
To calculate the molarity, first convert the mass of sodium nitrate to moles using its molar mass. Then, divide the number of moles by the volume of solution in liters (265 mL = 0.265 L) to get the molarity. The molarity of the solution is about 0.68 M.
To find the number of moles of sodium nitrate (NaNO3) in a 650 mL solution with a concentration of 0.28 M, use the formula: moles = concentration (M) × volume (L). First, convert 650 mL to liters: 650 mL = 0.650 L. Then, calculate the moles: 0.28 M × 0.650 L = 0.182 moles of NaNO3.
What is the volume of 35.7g of sodium chloride in 100cm3 of cold water?
Dilute 1 mL of 0.5 M silver nitrate solution to a total volume of 1 L with water to make a 1 mM silver nitrate solution.
The density of sodium at room temperature is 0.968g/cm3. Density = mass/volume. To find the volume of a substance from its density and mass, divide the mass by the density. V = M/D = 30g/0.968g/cm3 = 31cm3
Density=Mass/Volume