Partial vapour pressure of volatile liquids
Choose the right answer from following. For a solution of volatile liquids the partial vapour pressure of each component in solution is directly proportional to: (a) Molarity (b) Mole fraction (c) Molality (d) Normality
Two tanks which contain water are connected to each other through a valve. The initial conditions are as shown (at equilibrium):
Atomic orbitals can be combined, in a process called hybridization, to describe the bonding in polyatomic molecules. Descriptions of the bonding in CH4 can be used to illustrate the valence bond procedure. We must arrive a
The solution of sugar in water comprises: (i) Free atoms (ii) Free ions (iii) Free molecules (iv) Free atom and molecules. Choose the right answer from the above.
Choose the right answer from following. Vapour pressure of a solution is: (a) Directly proportional to the mole fraction of the solvent (b) Inversely proportional to the mole fraction of the solute (c) Inversely proportional to the mole fraction of the solvent (d
The coordination number of a cation engaging a tetrahedral hole is: (a) 6 (b) 8 (c) 12 (d) 4 Answer: (d) The co-ordination number of a cation occupying a tetrahedral hole is 4.
The accuracy of your written English will be taken into account in marking. 1. (a) Identify the spectator ions in the following equation &nb
Answer the following qustion. The definition “The mass of a gas dissolved in a particular mass of a solvent at any temperature is proportional to the pressure of gas over the solvent” is: (i) Dalton’s Law of Parti
Can someone please help me in getting through this problem. The normality of a solution of sodium hydroxide 100 ml of which includes 4 grams of NaOH is: (a) 0.1 (b) 40 (c) 1.0 (d) 0.4
Choose the right answer from following. Molar concentration (M) of any solution : a) No. of moles of solute/Volume of solution in litre (b) No. of gram equivalent of solute / volume of solution in litre (c) No. of moles os solute/ Mass of solvent in kg (
How to obtain relation between Vm and Km,given k(sec^-1) = Vmax/mg of enzyme x molecular weight x 1min/60 sec S* = 4.576(log K -10.753-logT+Ea/4.576T).
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