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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.
The volume of 24 mmol of ethyl iodide is 1,929 mL.
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 answer is 252,05 mL.
12.32 l n20
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.
The volume of 24 mmol of ethyl iodide is 1,929 mL.
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
What is the volume of 35.7g of sodium chloride in 100cm3 of cold water?
First, you must either find or be provided with a known mass of sodium hydroxide and a volumetric vessel. You must also know the molecular mass of sodium hydroxide, which is about 40.00. If the volume of sodium is sufficient, you can complete the preparation by determining the volume of the volume of the vessel in litres, multiply this volume by twice the molecular mass of sodium hydroxide, and dissolve the resulting mass in the known volume.
The volume occupied by 3 moles of nitrogen gas will be different, depending on the temperature and pressure of the gas.
Density=Mass/Volume
The answer is 252,05 mL.
The answer is 48,17 L.
This depends on: - the volume of the drop - the concentration of sodium chloride solution
0.5 ml or 1/2