- Dissolve 502 g uranyl nitrate hexahydrate in distilled water in A Berzelius flask.
- Transfer quantitatively the solution in a 1 L volumetric flask.
- Add distilled water to the mark, maintaining the flask in a thermostat, at 20 0C, for 30 min.
- Stir vigorously.
- Extract an aliquot for analysis, to know exactly the uranium concentration.
Yes. One mole of anything contains 6.02x10^23 "particles". In the case of the element uranium, it would be 6.02x10^23 atoms of uranium in 1 mole. In the case of CO2, it would be 6.02x10^23 molecules of CO2 in 1 mole.
The formula is C6H12O6 which is 180g/mole. Divide that in half for 90g in one liter of water for a 0.5 molar solution
Get moles by; Molarity = moles of solute/Liters of solution 0.50 M KNO3 = moles KNO3/2.0 L = 1.0 mole KNO3 Now find grams of 1.0 mole KNO3 1.0 mole KNO3 (101.11 grams/1 mole KNO3) = 101.11 grams KNO3 needed call it 100 grams
The lowest freezing point is observed for 1 mole of KOH, because its one moles produce 2 moles of ions in solution, 1 mole of cation K+ and 1 mole of anion OH-.
1.00 m
Dissolve 196,2 g substance in 1 L ethanol.
Yes. One mole of anything contains 6.02x10^23 "particles". In the case of the element uranium, it would be 6.02x10^23 atoms of uranium in 1 mole. In the case of CO2, it would be 6.02x10^23 molecules of CO2 in 1 mole.
9.6 grams of uranium (1 mole U/238.0 grams)(6.022 X 10^23/1 mole U) = 2.4 X 10^22 atoms of uranium
1 hydrogen mole = 1,007 94 g 1 uranium mole = 238,028 91 g But the number of atoms in a mole is the same for all the elements (Avogadro constant = 6,022 141 79(30) × 1023).
The formula is C6H12O6 which is 180g/mole. Divide that in half for 90g in one liter of water for a 0.5 molar solution
1 atomgram of uranium = 238,02891 gramsAnswer:The molar mass of Uranium is 238.03 g/mol
Approx. 12,5 cm3.
3.58 X 1024 atoms of uranium (1 mole U/6.022 X 1023) = 5.94 moles uranium to be precise and with significant figures -------------------------------------------------------------------------------
Dissolve 0.01 mole of KCl in upto 1 Liter. 0.01 mole = 0.01 (mol) * MKCl (g/mol KCl) = (0.01*MKCl) gram KCl per Liter
- calculate the molar mass of the compound from the atomic weights of the elements - exactly weight this mass on a laboratory balance - put the compound in an 1 L volumetric flask - dissolve the compound in distilled (deionized) water at 20 0C; solution up to the mark - stir vigorously the solution
Get moles by; Molarity = moles of solute/Liters of solution 0.50 M KNO3 = moles KNO3/2.0 L = 1.0 mole KNO3 Now find grams of 1.0 mole KNO3 1.0 mole KNO3 (101.11 grams/1 mole KNO3) = 101.11 grams KNO3 needed call it 100 grams
The lowest freezing point is observed for 1 mole of KOH, because its one moles produce 2 moles of ions in solution, 1 mole of cation K+ and 1 mole of anion OH-.