Which of the following functional groups comprises a carbon atom bonded to a hydroxyl group?
2 answers:
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The question is incomplete, here is the complete question.
If we start with 1.000 g of strontium-90, 0.805 g will remain after 9.00 yr. This means that the half-life of strontium-90 is ________ yr.
a. 28.8 b. 30.9 c. 35.4 d. 32.2
Answer : The half-life of strontium-90 is 28.8 years.
Explanation :
This is a type of radioactive decay and all radioactive decays follow first order kinetics.
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by :
where,
k = rate constant
t = time taken for decay process = 9.00 year
a = initial amount or moles of the reactant = 1.000 g
a - x = amount or moles left after decay process = 0.805 g
Putting values in above equation, we get:
To calculate the half-life, we use the formula :
Therefore, the half-life of strontium-90 is 28.8 years.
Answer and Explanation:
ΔH = (Bond energies of the products) - (Bond Energies of the reactants)
a) C2H6 + H2 ----> 2CH4 ΔH = -65.07 KJ/mol
Bond energy of product = 2×Bond energy of CH4 =2 × (4 (C-H)) =2 × 4 × 415 = 3320 KJ/mol
Bond energy of reactants = Bond energy of C2H6 + Bond energy of H2
Bond energy of C2H6 = (C-C) + 6(C-H) = 347 + 6(C-H)
Bond energy of H2 = (H-H) = 432 KJ/mol
-65.07 = 3320 - (Bond energy of C2H6 + 432)
Bond energy of C2H6 = 3320-432+65.07 = 2953.07 KJ/mol
Bond energy of C2H6 = (C-C) + 6(C-H) = 347 + 6(C-H) = 2953.07
(C-H) = (2953.07 -347)/6 = 434.345 KJ/mol
b) C2H4 + 2H2 ------> CH4 ΔH = -202.21 KJ/mol
Bond energy of product = 2×Bond energy of CH4 =2 × (4 (C-H)) =2 × 4 × 415 = 3320 KJ/mol
Bond energy of reactants = Bond energy of C2H4 + (2 × Bond energy of H2)
Bond energy of C2H4 = (C=C) + 4(C-H) = 614 + 4(C-H)
(2 × Bond energy of H2) = 2 × 432 = 864 KJ/mol
-202.21 = 3320 - (Bond energy of C2H4 + 864)
Bond energy of C2H4 = 3320+202.21-864 = 2658.1 KJ/mol
Bond energy of C2H4 = (C=C) + 4(C-H) = 614 + 4(C-H) = 2658.1
(C-H) = (2658.1 - 614)/4 = 511.05 KJ/mol
c) C2H2 + 3H2 ------> 2CH4 ΔH = -376.74 KJ/mol
Bond energy of product = 2×Bond energy of CH4 =2 × (4 (C-H)) =2 × 4 × 415 = 3320 KJ/mol
Bond energy of reactants = Bond energy of C2H2 + (3 × Bond energy of H2)
Bond energy of C2H = (triple bond Carbon to Carbon) + 2(C-H) = 839 + 2(C-H)
(3 × Bond energy of H2) = 3 × 432 = 1296 KJ/mol
-376.74 = 3320 - (Bond energy of C2H2 + 1296)
Bond energy of C2H2 = 3320+376.74-1296 = 2400.74 KJ/mol
Bond energy of C2H2 = (triple bond Carbon to Carbon) + 2(C-H) = 839 + 2(C-H) = 2400.74
(C-H) = (2400.74 - 839)/2 = 780.87 KJ/mol
QED!!!
