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NCERT
1 day, 4 hours ago

16. Why is that once a person starts taking alcohol or drugs, it is difficult to get rid of this habit. Discuss it with your teacher.

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S Sonika
Answered 1 day, 3 hours ago
Once a person starts taking alcohols or drug, it becomes very difficult to get rid of this habit. It is mainly because according to this person, alcohol is the only way to attain a normal state. Alcohol does not affect any organ other than the nervous system of a human body. However, due to the effect on the nervous system other organs also get affected. Prolonged use of alcohol can make a...
Engineering
6 days, 4 hours ago

Let \alpha and \beta   be the roots of equation

x^{2}-6x-2=0.\; if a_{n}=\alpha ^{n}-\beta ^{n},\; for\: n\geq 1,\; then\; the \: value\; of\; \frac{a_{10}-2a_{8}}{2a_{9}}

is equal to:

  • Option 1)

    6

  • Option 2)

    -6

  • Option 3)

    3

  • Option 4)

    -3

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S solutionqc
Answered 5 days, 4 hours ago
\(\\x^2-6x-2=0\\\alpha+\beta=6\\\alpha\cdot\beta=-2\\Given,\;\;\frac{a_{10}-2a_8}{2a_9},\;\;where\;\;a_n=\alpha^n-\beta^n\\\Rightarrow \frac{a_{10}-2a_8}{2a_9}=\frac{\alpha ^{10}-\beta ^{10}-2\left(\alpha \:^8-\beta \:^8\right)}{2\left(\alpha \:^9-\beta \:^9\right)}\\replace\;\;-2\;\;with\;\;\alpha\beta\\\Rightarrow \frac{\alpha ^{10}-\beta ^{10}+\left(\alpha \beta \right)\left(\alpha ^8-\beta...
NCERT
1 week, 1 day ago

Q.12.    What is the difference between the manner in which movement takes place in a sensitive plant and the movement in our legs?
 

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S seema garhwal
Answered 1 week ago
    Movement in sensitive plant Movement in human leg Movement in sensitive plants is due to involuntary action. It is in response to a stimulus (touch), for example, Mimosa Pudica movement in the human legs is a voluntary action no special tissues are needed for the transfer of information. The information is transferred through the nervous system.  
NCERT
1 week, 1 day ago

Q 15)
f(x)\left\{\begin{matrix} 2x & if &x<0 \\ 0& if &0\leq x\leq 1 \\ 4x&if & x>1 \end{matrix}\right.

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G Gautam harsolia
Answered 1 week, 1 day ago
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...
NCERT
1 week, 6 days ago

4.30   The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

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M manish
Answered 1 week, 2 days ago
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
NCERT
1 week, 6 days ago

4.29   The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 \times 10 ^{10} s ^{-1} . Calculate k at 318K and Ea.

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M manish
Answered 1 week, 2 days ago
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 =...
NCERT
1 week, 6 days ago

4.27   The rate constant for the first order decomposition of H_2O_2 is given by the following equation:
           \log k = 14.34 - 1.25 \times 10 ^ 4K/T .

Calculate E_a for this reaction and at what temperature will its half-period be 256 minutes?

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M manish
Answered 1 week, 2 days ago
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)    
NCERT
1 week, 6 days ago

4.26   The decomposition of hydrocarbon follows the equation  k = (4.5 × 1011s^{-1}) e^{-28000K/T}. Calculate E_{a}

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M manish
Answered 1 week, 2 days ago
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
NCERT
1 week, 6 days ago

4.25   Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t _{1/2 } = 3.00 hours. What fraction of sample of sucrose remains after 8 hours ?

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M manish
Answered 1 week, 2 days ago
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
NCERT
1 week, 6 days ago

4.23   The rate constant for the decomposition of hydrocarbons is 2.418 \times 10 ^{-5} s ^{-1} at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

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M manish
Answered 1 week, 2 days ago
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)
NCERT
1 week, 6 days ago

4.22   The rate constant for the decomposition of N2O5 at various temperatures
             is given below:

           

            Draw a graph between ln k and 1/T and calculate the values of A and
            E_a. Predict the rate constant at 30° and 50°C.

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M manish
Answered 1 week, 2 days ago
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
NCERT
1 week, 6 days ago

4.21   The following data were obtained during the first order thermal decomposition of SO_2 Cl_2 at a constant volume.
            ( SO_2Cl_2 g) \rightarrow SO_2 (g) + Cl_2 (g)

        

           Calculate the rate of the reaction when total pressure is 0.65 atm.

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M manish
Answered 1 week, 2 days ago
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...
NCERT
1 week, 6 days ago

4.20   For the decomposition of azoisopropane to hexane and nitrogen at 543K, the following data are obtained.

            

          Calculate the rate constant.

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M manish
Answered 1 week, 2 days ago
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,                    
NCERT
1 week, 6 days ago

4.19   A first order reaction takes 40 min for 30% decomposition. Calculate t_{1/2}

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M manish
Answered 1 week, 2 days ago
For the first-order reaction,               (30% already decomposed and remaining is 70%)      therefore half life = 0.693/k                             =                              = 77.7 (approx) 
NCERT
1 week, 6 days ago

4.18   For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

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M manish
Answered 1 week, 2 days ago
case 1- for 99% complition,             CASE- II for 90% complition,            Hence proved.
NCERT
1 week, 6 days ago

4.17   During nuclear explosion, one of the products is ^{90} Sr with half-life of 28.1 years. If 1 \mu g of ^{90}Sr  was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

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M manish
Answered 1 week, 2 days ago
Given, half life = 21.8 years            = 0.693/21.8             and,  by putting the value we get,                            taking antilog on both sides, [R] = antilog(-0.1071)       = 0.781  Thus 0.781  of  will remain after given 10 years of time. Again,  Thus 0.2278  of  will remain after 60 years.               
NCERT
1 week, 6 days ago

4.16   The rate constant for a first order reaction is 60 s^{-1}. How much time will it take to reduce the initial concentration of the reactant to its 1/16th  value?

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M manish
Answered 1 week, 2 days ago
We know that, for first order reaction,           (nearly) Hence the time required is  
NCERT
1 week, 6 days ago

4.15 (6)   The experimental data for decomposition of   N_2O_5 [2N_2O_5 \rightarrow 4NO_2 + O_2]  in gas phase at 318K are given below:

                

              Calculate the half-life period from k and compare it with (ii).

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M manish
Answered 1 week, 2 days ago
The half life produce =                                    =                                      
NCERT
1 week, 6 days ago

4.15 (5)   The experimental data for decomposition of N_2O_5 [2N_2O_5 \rightarrow 4NO_2 + O_2] in gas phase at 318K are given below:

                 

                Calculate the rate constant.
                

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M manish
Answered 1 week, 2 days ago
From the log graph, the slope of the graph is =                                            = -k/2.303                             ..(from log equation)        On comparing both the equation we get,           
NCERT
1 week, 6 days ago

4.15 (4)    The experimental data for decomposition of  N_2O_5 [2N_2O_5 \rightarrow 4NO_2 + O_2] in gas phase at 318K are given below:

              

                What is the rate law ?

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M manish
Answered 1 week, 2 days ago
Here, the reaction is in first order reaction because its log graph is linear. Thus rate law can be expessed  as 
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