An inexpensive and accurate method of measuring the quantity of electricity passing through a circuit is to pass it through a solution of a metal ion and weigh the metal deposited. A silver electrode immersed in an Ag solution weighs 1.7854 g before the current passes and weighs 1.9274 g after the current has passed. how many coulombus have passed?

127.013 C

Explanation:

-We apply Faraday's First Law of Electrolysis which states that the mass,m of a substance deposited is directly proportional to the current,Q passed through it:

Where Z is the proportionality constant(g/C)

-The current can therefore be calculated as:

Hence, the amount of current through the circuit is 127.013 C

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

pla·gia·rism

/ˈplājəˌrizəm/

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noun

the practice of taking someone else's work or ideas and passing them off as one's own.

explanation: