The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is -5084.1 kJ. You may want to reference (Pages 381 - 385) Section 9.6 while completing this problem.

This is an incomplete question, here is a complete question.

The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is -5084.1 kJ. You may want to reference (Pages 381 - 385) Section 9.6 while completing this problem. If the change in enthalpy is -5074.2 kJ, how much work is done during the combustion? Express the work in kilojoules to three significant figures.

Answer : The work done during the combustion is, 9.9 kJ

Explanation :

Formula used :

where,

w = work done = ?

= change in enthalpy  = -5074.2 kJ

= change in internal energy  = -5084.1 kJ

R = gas constant = 8.314 J/mol.K

Now put all the given values in the above formula, we get:

Thus, the work done during the combustion is, 9.9 kJ

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

b. examine and describe cells