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Half life of the reaction is-
time corresponding to the $1.63 \times 10^{2}/2$ mol/ L  = 81.5 mol /L is the half life of the reaction.  From the graph, the answer shoul be in the range of 1400 s to 1500 s.

On increasing time, the concentration of  gradually decreasing exponentially.
Given , half-life of radioactive decay = 5730 years So,                  per year we know that, for first-order reaction,                  = 1845 years (approximately) Thus, the age of the sample is 1845 years
The half-life for the first-order reaction is                                                                               = 0.693/4                                                                             =  0.173 year (approximately)
the half-life for the first-order reaction is expressed as ;           =  0.693/2           = 0.35 min (approx)
We know that, half-life () for first-order reaction =                                                             =
The given reaction is first order wrt A and zero order in wrt B. So, the rate of reaction can be expressed as;   Rate = k[A] from exp 1, k = 0.2 per min. from experiment 2nd, [A] =  from experiment 3rd,             from the experiment 4th, from here [A] = 0.1 mol/L
Let assume the rate of reaction wrt A is  and wrt B is  . So, the rate of reaction is expressed as- Rate =  According to given data,  these are the equation 1, 2, 3 and 4 respectively Now,  divide the equation(iv) by (i) we get, from here we calculate that  Again, divide equation (iii) by (ii) from here we can calculate the value of y is 2 Thus, the rate law is now,   So,         ...
we know that  rate law () =  As per data   these are the equation 1, 2 and 3 respectively Now, divide eq.1 by equation2, we get from here we calculate that y = 0 Again, divide eq. 2 by Eq. 3, we get Since y =0 also substitute the value of y  So,   =  =  taking log both side we get,     = 1.496     = approx 1.5 Hence the order of reaction w.r.t A is 1.5 and w.r.t B is 0(zero)

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(v)12-16 and 13-15 group compounds 12-16 group compounds: Compounds formed between elements of group 12 and elements of group 16 are called 12-16 group compounds. Example: ZnS 13-15 group compounds: Compounds formed between elements of group 13 and elements of group 15 are called 13-15 group compounds. Example: GaAs

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(iv) Antiferromagnetism :  Substances like MnO showing antiferromagnetism have domain structure similar to ferromagnetic substance, but their domains are oppositely oriented and cancel out each other's magnetic moment as shown in the figure.

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(iii)    Ferrimagnetism :  When the magnetic moments of the domains in the substance are aligned in parallel and anti-parallel directions in unequal numbers than ferrimagnetism is observed. Refer to the given figure. They are weakly attracted by a magnetic field as compared to ferromagnetic substances. Example:. These substances also lose ferrimagnetism on heating and become paramagnetic.

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(ii) Paramagnetism :  The substances that are weakly attracted by a magnetic field are called paramagnetic substances. They are magnetised in a magnetic field in the same direction. They lose their magnetism in the absence of a magnetic field. Paramagnetism is due to the presence of one or more unpaired electrons which are attracted by the magnetic field. Example:.

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(i) Ferromagnetism Substances that are attracted very strongly by a magnetic field are called ferromagnetic substances. Example: iron, cobalt, nickel, gadolinium and CrO2. Besides strong attractions, these substances can be permanently magnetised. The metal ions of ferromagnetic substances are grouped together into small regions called domains. Thus, each domain acts as a tiny magnet. In an...

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NaCl is doped with 10–3 mol % of  Concentration in % so that take a total of 100 mol of solution. Moles of NaCl = 100 - moles of  Moles of  is very less , so we can neglect them. Moles of NaCl =100 1 mole of NaCl is dipped with =   mol of . So cation vacanties per mole of NaCl = mol   particles  So cation vacancies per mol of NaCl =                                                         ...

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For cubic close-packed structure, we have  Given :                          the volume of one unit cell                                                                                    Total unit cells in 1.00 cm3

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For  cubic close-packed structure, we have  Given :

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When any negative ion is absent from lattice site then result the crystal now has an excess of cations. To maintain electrical neutrality the vacant anionic site is occupied by an electron as shown in the figure. The anionic sites occupied by unpaired electrons are called F-centres. The F- centre is responsible for most interstitial properties of the compound. This defect can be observed in  NaCl.