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 shows Frenkel defect because of the small size of  ion, it can easily fit in the interstitial sites while  can not as it is bigger in size.

A binary number uses only two digits zero and one. In a binary system, numbers are represented as follows

NUMBER AS INTEGER NUMBER IN BINARY
0 00
1 01
2 10
3 11
4 100
5 101
6 110
7 111

 

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Ever observed an old glass window? It will be thicker at the base than the top.                        This shows that glass flows(at a very slow rate) and it can flow because it is amorphous.

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It's called dependence and addiction. If the use or intake of alcohol is suddenly discontinued it leads to withdrawal symptoms.

 

Given function is Given function is satisfies for the all real values of x case (i)  k < 0 Hence, function is continuous for all values of x < 0 case (ii)  x = 0 L.H.L at x= 0 R.H.L. at x = 0 L.H.L. = R.H.L. = f(0) Hence, function is continuous at x = 0 case (iii)  k > 0 Hence , function is continuous for all values of x > 0 case (iv) k < 1  Hence , function is...
From the Arrhenius equation, ...................................(i) it is given that  T1= 293 K T2 = 313 K Putting all these values in equation (i) we get, Activation Energy = 52.86 KJ/mol   This is the required activation energy
We know that, for a first order reaction- Case 1 At temp. = 298 K    = 0.1054/k Case 2 At temp = 308 K       = 2.2877/k' As per the question    K'/K = 2.7296 From Arrhenius equation,      = 76640.096 J /mol      =76.64 KJ/mol   k at 318 K we have , T =318K                  A=  Now  After putting the calue of given variable, we get  on takingantilog we get,  k =...
The decomposition of A into a product has a value of k as at 10°C and energy of activation 60 kJ mol–1. K1 =  K2 =   =  60 kJ mol–1 K2 = 
The Arrhenius equation is given by  taking log on both sides,  ....................(i) given equation, .....................(ii) On comparing both equation we get, activation energy  half life ()  = 256 min k = 0.693/256    With the help of equation (ii),            T =                 = 669 (approx)    
The Arrhenius equation is given by  .................................(i) given equation, ............................(ii) by comparing equation (i) & (ii) we get, A= 4.51011 per sec Activation energy = 28000  (R = 8.314)                              = 232.792 KJ/mol
For first order reaction, given that half life = 3 hrs () Therefore k = 0.693/half-life                     = 0.231 per hour Now,                   = antilog (0.8024)                   = 6.3445 (approx) Therefore fraction of sample of sucrose remains after 8 hrs is 0.157
Given that, k =  t = 100 s Here the unit of k is in per sec, it means it is a first-order reaction. therefore,  Hence the concentration of rest test sample is 0.135 mol/L
Given that, k =  = 179.9 KJ/mol T(temp) = 546K According to Arrhenius equation, taking log on both sides,                             = (0.3835 - 5) + 17.2082               = 12.5917 Thus A = antilog (12.5917)          A = 3.9  per sec (approx)
From the above data, T/ 0 20 40 60 80 T/K 273 293     313 333 353 () 3.66 3.41 3.19 3.0 2.83 0.0787 1.70 25.7 178 2140 -7.147 -4.075 -1.359 -0.577 3.063 Slope of line  =  According to Arrhenius equations, Slope =    12.30 8.314             = 102.27  Again, When T = 30 +273 = 303 K and 1/T =0.0033K    k =  When T = 50  + 273 = 323 K  and 1/T = 3.1  K k = 0.607 per sec
The thermal decomposition of  is shown here; After t time, the total pressure  =                                                 So,  thus,  for first order reaction,     now putting the values of pressures, when t = 100s       when      = 0.65 - 0.5    = 0.15 atm So,                              = 0.5 - 0.15                             = 0.35 atm Thus, rate of reaction, when the...
Decompostion is represented by equation- After t time, the total pressure  =                                                 So,  thus,  for first order reaction,     now putting the values of pressures, when t =360sec when t = 270sec So,                    
For the first-order reaction,               (30% already decomposed and remaining is 70%)      therefore half life = 0.693/k                             =                              = 77.7 (approx) 
case 1- for 99% complition,             CASE- II for 90% complition,            Hence proved.
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