In part b the given conditions were 1.00 mol of argon in a 0.500-l container at 18.0°c. you identified that the ideal pressure (p ) was 47.8 atm, and the real pressure (prral) was 45.7 atm under these conditions. complete the sentences to analyze this difference match the words and compounds in the left column to the appropriate blanks in the sentences on the right. make certain each sentence is complete before submitting your answer. view available hint(s) reset strong intermolecular forces the percent difference between the ideal and real gas is decrease this difference is considered significant, and is best explained because argon atoms have relatively small molecular volumes if the volume of the container were increased to 2.00 l, you would expect the percent difference between the ideal and real gas to 4.38% 4.48% 4.58% increase weak intermolecular forces large molecular volumes

4,38%

small molecular volumes

Decrease

Explanation:

The percent difference between the ideal and real gas is:

(47,8atm - 45,7 atm) / 47,8 atm × 100 = 4,39% ≈ 4,38%

This difference is considered significant, and is best explained because argon atoms have relatively small molecular volumes. That produce an increasing in intermolecular forces deviating the system of ideal gas behavior.

Therefore, an increasing in volume will produce an ideal gas behavior. Thus:

If the volume of the container were increased to 2.00 L, you would expect the percent difference between the ideal and real gas to decrease

I hope it helps!

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

answer; protein;