Spectroscopy homework
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Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid, establishes a buffer with a pH of approximately 5.
Select the right answer of the question. Which is heaviest: (a)25 gm of mercury (b)2 moles of water (c)2 moles of carbon dioxide (d)4 gm atoms of oxygen
Partition function; that the translational energy of 1 mol of molecules is 3/2 RT will come as no surprise. But the calculation of this result further illustrates the use of quantized states and the partition function to obtain macroscopic properties. The partition fu
Relative humidity is the ratio of the partial pressure of water in air to the partial pressure of water in air saturated with water at the same temperature, stated as a percentage: Relative = Q : Question based on relative lowering of Give me answer of this question. When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to: (a) Mole fraction of solute (b) Mole fraction of solvent (c) Concentration of the solute in grams per litre (d) Concentratio
Give me answer of this question. When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to: (a) Mole fraction of solute (b) Mole fraction of solvent (c) Concentration of the solute in grams per litre (d) Concentratio
why BiF3 is ionic whereas other trihalides are covalent in nature?
Transference numbers and molar conductors can be used to calculate ionic mobilities. This tables under is giving the transference numbers for positive ions at 25 degree C and the values obtained by extrapolation to infinite dilution: Q : Product of HCl Zn Illustrate the Illustrate the product of HCl Zn?
Illustrate the product of HCl Zn?
Choose the right answer from following. If 0.50 mol of CaCl2 is mixed with 0.20 mol of Na3PO4, the maximum number of moles of Ca3 (PO2)2 which can be formed: (a) 0.70 (b) 0.50 (c) 0.20 (d) 0.10
The free energy of a component of a liquid solution is equal to its free energy in the equilibrium vapour.Partial molal free energies let us deal with the free energy of the components of a solution. We use these free energies, or simpler concentration ter
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