Carbon and oxygen combine to form two oxides having the composition : 1st oxide C = 42.9% and 2nd oxide C = 27.3% . Show that the data is in agreement with the law of multiples proportions.

Weight of copper oxide obtained by treating 2.16 g metallic copper with nitric acid and subsequent ignition was 2.70 g. In another experiment, 1.15 g of copper oxide on reduction yielded 0.92 g of copper. Show that these results illustrate the law of definite proportions.

A non-metal A which is the largest consituent of air, when heated with H_2 in 1:3 ratio in the presence of catalyst gives a gas B. on heating with O_2 it gives an oxide C. If this oxide is passed into water in the presence of air it gives an acid D which acts as a strong oxidising agent. To which group of periodic table does this non-metal belong?

If hydrogen atoms (in the ground state ) are passed through an homogeneous magnetic field, the beam is split into two parts. This interaction with the magnetic field shows that the atoms must have magnetic moment. However, the moment cannot be due to the orbital angular momentum since l=0. Hence one must assume existence of intrinsic angular momentum, which as the experiment shows, has only two permitted orientations. Spin of the electron produce angular momentum equal to S=sqrt(s(s+1))(h)/(2pi) where S=+(1)/(2) . Total spin of an atom = +(n)/(2) " or "-(n)/(2) where n is the number of unpaired electrons. The substance which contain species with unpaired electrons in their orbitals behave as paramagnetic substances. The paramagnetism is expressed in terms of magnetic moment. The magnetic moment of an atom mu_(s)sqrt(s(s+1))(eh)/(2pimc)=sqrt((n)/(2)((n)/(2)+1))(eh)/(2pimc)" "s=(n)/(2) impliesmu_(s)=sqrt(n(n+1)) B.M. 1. B.M. (Bohr magneton)= (eh)/(4pimc) If magnetic moment is zero the substance is diamagnetic. If an ion of _(25)Mn has a magnetic moment of 3.873 B.M. Then oxidation state of Mn in ion is :

A sample of calcuium carbonate (CaCO_(3)) has the following percentage composition: Ca=40%, C=12%, O=48% If the law of constant proportions is true. Then the weight of calcium in 4 g of a sample of calcium carbonate obtained from another source will be

A non-metal A which is the largest consituent of air, when heated with H_2 in 1:3 ratio in the presence of catalyst gives a gas B. on heating with O_2 it gives an oxide C. If this oxide is passed into water in the presence of air it gives an acid D which acts as a strong oxidising agent. Identify A,B,C and D.

Two reactions of the same order have equal pre-exponential factors but their activation energies difter by 24.9 kJ mol^-1 . Calculate the ratio between the rate constants of these reactions at 27^@C . (Gas constant, R= 8.3 JK^-1 mol^-1 ).

There is another useful system of units, besides the SI/MKS. A system, called the cgs system. In this system Coulumb's law is given by vecF = (Qq)/(r^2) hatr where the distance r is measured in cm (=10^(-2) m) , F in dynes (= 10^(-5) N) and the charges in electrostatic units (es units), where 1es unit of charge = 1/[3] xx 10^(-9) C . The number [3] actually arises the speed of light in vacuum which now taken to be exactly given by c = 2.99792458 xx 10^8 m/s . An approximate value of c then is c = [3] xx 10^8 m/s . Show that the Coulomb law in cgs units yields. 1 esu of charge = 1 (dy )^(1/2) cm . obtain the dimension of units of charge in terms of mass M, length L and time T. show that it is given in terms of fractional powers of M and L.

A part of Ellingham diagram for some metal oxides (based upon 1 mole of O_2 ) and carbon is shown. In figure A, B, C and D represent curves for metal oxides and a, b, c, d, e and f are temperatures. Answer the following : Will B oxide reduce metal oxide of A or C or both ?

The strength of H_(2)O_(2) is expressed in several ways like molarity, normality,% (w/V), volume strength, etc. The strength of "10 V" means 1 volume of H_(2)O_(2) on decomposition gives 10 volumes of oxygen at 1 atm and 273 K or 1 litre of H_(2)O_(2) gives 10 litre of O_(2) at 1 atm and 273 K The decomposition of H_(2)O_(2) is shown as under : H_(2)O_(2)(aq) to H_(2)O(l)+(1)/(2)O_(2)(g) H_(2)O_(2) can acts as oxidising as well as reducing agent. As oxidizing agent H_(2)O_(2) is converted into H_(2)O and as reducing agent H_(2)O_(2) is converted into O_(2) . For both cases its n-factor is 2. :. "Normality " "of" H_(2)O_(2) " solution " =2xx " molarity of" H_(2)O_(2) solution What is the molarity of "11.2 V" H_(2)O_(2) ?

9 moles of "D" and 14 moles of E are allowed to react in a closed vessel according to given reactions. Calculate number of moles of B formed in the end of reaction, if 4 moles of G are present in reaction vessel. (percentage yield of reaction is mentioned in the reaction) Step -1 3D+4E 80%rarr5C+A Step-2 3C+5G 50%rarr 6B+F