A50 ml beaker has an uncertainty of ± 10%. what is the absolute uncertainty for this beaker?

The absolute uncertainty is 10%.

Explanation:

The absolute uncertainty, sometimes referred to as absolute error, is the size of the possible range of values where by the actual or true value of a measurement probably lies.

The uncertainty of the 50 mL beaker is given as  ± 10%. This implies that the volume recorded using the beaker will either be 10 % above or below the actual volume.

The absolute uncertainty ignores the direction of the error, whether above or below, and is mostly concerned with the magnitude of the error. We simply ignore the signs of the uncertainty and this will give the value of the absolute uncertainty.

The absolute uncertainty for this beaker is ± 5 mL.

Explanation

The uncertainty is the measure of error an instrument can exhibit during measurement.

It is the error range in any kind of measurement.

There are two kinds of uncertainty: relative and absolute.

Relative uncertainty is the percentage of error occurred during measurement while absolute uncertainty quantifies the perfect value of error range with the measuring unit.

In this problem, the relative uncertainty was given as ±10 % and the estimated value is 50 mL.

So the conversion from relative uncertainty to absolute uncertainty can be done by the following formula

Absolute uncertainty=Relative uncertainty*Estimated value

Absolute uncertainty=10/100*50=5 mL

So the value can be written as 50 mL±5 mL with the absolute uncertainty as 5 mL.

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

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.