chem
Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid.
171 g of cane sugar (C12H22O11) is dissolved in one litre of water. Find the molarity of the solution: (i) 2.0 M (ii) 1.0 M (iii) 0.5 M (iv) 0.25 M Choose the right answer from above.
The surface between a liquid and a vapour distinguishes these fluids. The surface tension of liquids can be looked upon as that the property which draws a liquid together and forms a liquid vapour interface, therefore, distinguishing liquids from gases.<
what is molarity of a solution of hcl which contains 49% by weight of solute and whose specific gravity is 1.41
The integrated forms of the first order rate equations are conveniently used to compare concentration time results with this rate equation. Rate equations show the dependence of the rate of the reaction on concentration can be integrated to give expressions fo
The objective of this work is to separate a binary mixture and to cool down the bottom product for storage. (Check table below to see which mixture you are asked to study). 100 kmol of feed containing 10 mol percent of the lighter component enters a continuous distillation column at the m
Provide solution of this question. While 1.80gm glucose dissolve in 90 of H2O , the mole fraction of glucose is: (a) 0.00399 (b) 0.00199 (c) 0.0199 (d) 0.998
Colligative properties of a solution depends upon: (a) Nature of both solvent and solute (b) The relative number of solute and solvent particles (c) Nature of solute only (d) Nature of solvent only
Give me answer of following question. The sum of the mole fraction of the components of a solution is : (a) 0 (b) 1 (c) 2 (d) 4.
Can someone please help me in getting through this problem. 2.0 molar solution is acquired, when 0.5 mole solute is dissolved in: (i) 250 ml solvent (ii) 250 g solvent (iii) 250 ml solution (iv) 1000 ml solvent
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
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