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NCERT
2 days, 20 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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B Bipasha Kumar
Answered 15 hours, 23 minutes ago

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

 

Engineering
1 week 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 6 days, 20 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, 2 days 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, 2 days 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, 2 days 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, 2 days 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
2 weeks 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, 4 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
2 weeks 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, 4 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
2 weeks 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, 4 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
2 weeks 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, 4 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
2 weeks 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, 4 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
2 weeks 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, 4 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
2 weeks 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, 4 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
2 weeks 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, 4 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
2 weeks 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, 4 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
2 weeks 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, 4 days ago
For the first-order reaction,               (30% already decomposed and remaining is 70%)      therefore half life = 0.693/k                             =                              = 77.7 (approx) 
NCERT
2 weeks 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, 4 days ago
case 1- for 99% complition,             CASE- II for 90% complition,            Hence proved.
NCERT
2 weeks 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, 4 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
2 weeks 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, 4 days ago
We know that, for first order reaction,           (nearly) Hence the time required is  
NCERT
2 weeks 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, 4 days ago
The half life produce =                                    =                                      
NCERT
2 weeks 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, 4 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
2 weeks 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, 4 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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