Lab question
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.
Many enzyme catalyzed reactions obeys a complex rate equation that can be written as the total quantity of enzyme and the whole amount of substrate in the reaction system. Many rate equations that are more complex than first and se
what is molarity of a solution of hcl which contains 49% by weight of solute and whose specific gravity is 1.41
Give me answer of this question. The atmospheric pressure is sum of the: (a) Pressure of the biomolecules (b) Vapour pressure of atmospheric constituents (c) Vapour pressure of chemicals and vapour pressure of volatile (d) Pressure created on to atmospheric molecules
Ionic dissociation depends on the dielectric constant of the solvent.The Arrhenius that ions are in aqueous solutions in equilibrium with parent molecular species allows many of the properties of ionic solutions to be understood. But difficulties began to
A student was analyzing an unknown containing only Group IV cations. When the unknown was treated with 3M (NH4)2CO3 solution, a white precipitate formed. Because the acetic acid bottle was empty, the student used 6M HCl to dissolve the precipitate. Following the procedure of this experiment, the stu
From where the tin is obtained? Briefly illustrate it.
Can someone please help me in getting through this problem. The solution ofAl2(SO4)3 d = 1.253gm/m comprise 22% salt by weight. The molarity, normality and molality of the solution is: (1) 0.805 M, 4.83 N, 0.825 M (2)
Can someone please help me in getting through this problem. Concentrated H2SO4 has a density of 1.98 gm/ml and is 98% H2SO4 by weight. The normality is: (a) 2 N (b) 19.8 N (c) 39.6 N (d) 98
It may be prepared by the action of phosphorus on thionyl chloride.P4 + 8SOCl2 4
20 ml of HCL solution needs 19.85 ml of 0.01M NaOH solution for complete neutralization. Morality of the HCL solution is: (i) 0.0099 (ii) 0.099 (iii) 0.99 (iv) 9.9 Choose the right answer from above.
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