Aflat plug 30 mm thick having an area of 4.0 × 10–4m2 and made of vulcanized rubber is used for closing an opening in a container. the gas co2 at 25°c and 2.0 atm pressure is inside the container. calculate the total leakage or diffusion of co2 through the plug to the outside in kg mol co2/s at steady state. assume that the partial pressure of co2 outside is zero. from barrer (b1), the solubility of the co2 gas is 0.90 m3 gas (at stp of 0°c and 1 atm) per m3 rubber per atm pressure of co2. the diffusivity is 0.11 × 10–9m2/s.

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

possibly, yes.

explanation:

the amount of sugar you take in is dangerously high, risk of heart disease, high risk of diabetes, and etc.

i think its 100

i hope this !