A sports car requires 10 kg of 93 octane to fill the tank. How many grams of octane are in the tank when it is full?

There are 10,000 grams of octane in the tank of the sports car when it is full

Explanation:

To solve this problem, we simply need to convert the mass of octane- 93  from kilograms to grams, using the standard conversion factor which is given below.

it will take 10kg of octane-93 to  fill up the tank

1 kg = 1000grammes

therefore 10 kg = 10 X 1000 = 10,000 grams.

Hence, there are 10,000 grams of octane in the tank of the sports car when it is full.

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

1.43 x 10^-11 j the guy commented the awnser just making sure u see it .