Q81.Let ๐, ๐, ๐ be the three distinct positive real numbers such that 2๐log๐๐= ๐๐log๐๐ and ๐log๐2 = ๐log๐๐ Then 6๐+ 5๐๐ is equal to ______.
What This Question Tests
This question requires manipulating given logarithmic and exponential equations to solve for unknown variables and evaluate a final expression.
Concepts Tested
Formulas Used
log(a^b) = b log(a)
e^(ln x) = x
๐ NCERT Sections This Tests
3.26 โ The Decomposition Of Hydrocarbon Follows The Equation
Chemistry Class 11 ยท Chapter 3
3.26 The decomposition of hydrocarbon follows the equation k = (4.5 ร 1011sโ1) e-28000K/T Calculate Ea. 87 Chemical Kinetics Reprint 2025-26
3.10 โ In A Reaction Between A And B, The Initial Rate Of Reaction (R0) Was Measured
Chemistry Class 11 ยท Chapter 3
3.10 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below: A/ mol Lโ1 0.20 0.20 0.40 B/ mol Lโ1 0.30 0.10 0.05 r0/mol Lโ1sโ1 5.07 ร 10โ5 5.07 ร 10โ5 1.43 ร 10โ4 What is the order of the reaction with respect to A and B? 3.11 The following results have been obtained during the kinetic studies of the reaction: 2A + B ยฎ C + D Experiment [A]/mol Lโ1 [B]/mol Lโ1 Initial rate of formation of D/mol Lโ1 minโ1 I 0.1 0.1 6.0 ร 10โ3 II 0.3 0.2 7.2 ร 10โ2 III 0.3 0.4 2.88 ร 10โ1 IV 0.4 0.1 2.40 ร 10โ2 Determine the rate law and the rate constant for the reaction. 3.12 The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/ mol Lโ1 [B]/ mol Lโ1 Initial rate/ mol Lโ1 minโ1 I 0.1 0.1 2.0 ร 10โ2 II โ 0.2 4.0 ร 10โ2 III 0.4 0.4 โ IV โ 0.2 2.0 ร 10โ2 3.13 Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 sโ1 (ii) 2 minโ1 (iii) 4 yearsโ1 3.14 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample. 3.15 The experimental data for decomposition of N2O5 [2N2O5 ยฎ 4NO2 + O2] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 102 ร [N2O5]/ 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 mol Lโ1 (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N2O5] and t. (iv) What is the rate law ? Chemistry 86 Reprint 2025-26 (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).
3.23 โ The Rate Constant For The Decomposition Of Hydrocarbons Is 2.418 ร 10โ5Sโ1
Chemistry Class 11 ยท Chapter 3
3.23 The rate constant for the decomposition of hydrocarbons is 2.418 ร 10โ5sโ1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
๐ Question Details
- Chapter
- Logarithms
- Topic
- Properties of logarithms and exponents
- Year
- 2023
- Shift
- 10 Apr Shift 1
- Q Number
- Q81
- Type
- Numerical
- NCERT Ref
- Class 11 Mathematics Ch 3: Trigonometric Functions (introduces logs informally, formal log properties in higher math context)