Molarity of acid solution
If 20ml of 0.4N, NaoH solution completely neutralises 40ml of a dibasic acid. The molarity of the acid solution is: (a) 0.1M (b) 0.2M (c) 0.3M (d) 0.4M Choose the right answer fron above.
If 20ml of 0.4N, NaoH solution completely neutralises 40ml of a dibasic acid. The molarity of the acid solution is: (a) 0.1M (b) 0.2M (c) 0.3M (d) 0.4M
Choose the right answer fron above.
Provide solution of this question. If 25 ml of 0.25 M NaCl solution is diluted with water to a volume of 500ml the new concentration of the solution is : (a) 0.167 M (b) 0.0125 M (c) 0.833 M (d) 0.0167 M
Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal
The symmetry of the molecules can be described in terms of electrons of symmetry and the corresponding symmetry operations.Clearly some molecules, like H2O and CH4, are symmetric. Now w
The equilibrium constant can be treated as a particular type of molecular distribution. Consider the simplest gas-phase reaction, one in which molecules of A are converted to molecules of B. the reaction, described by the equation Q : Organic and inorganic substances living living beings are made up of organic and inorganic substances.according to their complexity of their molecules how can ach of these substances be classified?
living beings are made up of organic and inorganic substances.according to their complexity of their molecules how can ach of these substances be classified?
Help me to solve this Question. The number of moles of SO2Cl2 in 13.5 gm is in is : (a) 0.1 (b) 0.2 (c) 0.3 (d) 0.4
Detail the moles of HCl which are present in .70 L of a .33 M HCl solution?
Help me to solve this problem. 0.5 M of H2AO4 is diluted from 1 lire to 10 litre, normality of resulting solution is : (a)1 N (b) 0.1 N (c)10 N (d)11 N
Give me answer of this question. The formula weight of H2SO4 is 98. The weight of the acid in 400mi of solution is: (a)2.45g (b) 3.92g (c) 4.90g (d) 9.8g
The first method begins with a well defined layer, or boundary, of solution near the center of rotation and tracks the movement of this layer to the outside of the cell as a function of time. Such a method is termed a sedimentary velocity experiment. A
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