Which liquids would cause the largest decrease in mass of a potato stick

distilled water I guess !

Just a guess though

If you find this useful, please mark my answer as the brainliest.

Explanation:

If you find this useful, please mark my answer as the brainliest.

distilled water

Explanation:

Some students investigated osmosis in raw potato sticks. The students measured the mass of three potato sticks using an electronic balance. The students left each potato stick in one of the three different liquids for 5 hours:i. distilled water. ii. dilute sodium chloride solution. iii. concentrated sodium chloride solution. After 5 hours they measured the mass again and calculated the change in mass. 1. Predict which of the liquids would cause the largest decrease in mass of a potato stick. 2. After the experiment, the students noticed that the potato stick with the lowest mass was soft and floppy. Explain why the potato stick had become soft and floppy. 3. The students followed the same experimental procedure with boiled potato sticks and found no overall change in mass in any of the solutions. Suggest why the mass of the boiled potato sticks remained the same.

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