1. (8pt) Using dimensional analysis convert 600.0 calories into kilojoules 2. (8pts) how much heat and kilojoules is needed to raise the temperature of 236 g of silver from 8.5°C to 34.9°C? (Cag = 0.24 J/g°C)

1. 2.510kJ  

2. Q = 1.5 kJ

Explanation:

Hello there!

In this case, according to the given information for this calorimetry problem, we can proceed as follows:

1. Here, we consider the following equivalence statement for converting from calories to joules and from joules to kilojoules:

Then, we perform the conversion as follows:

2. Here, we use the general heat equation:

And we plug in the given mass, specific heat and initial and final temperature to obtain:

Regards!

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

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