The radioisotope xenon-133 has a half-life of 5.2 days. How much of an 80 gram sample would be left after 20.8 days? PLEASE SHOW YOUR WORK

6.73g

Explanation:

T½ = 5.2days

No = 80g

N = ?

T = 20.8days

We'll have to find the disintegration constant first so that we can plug it into the equation that will help us find the mass of the sample after 20.8 days

T½ = In2 / λ

T½ = half life

λ = disintegration constant

λ = In2 / T½

λ = 0.693 / 5.8

λ = 0.119

In(N / No) = -λt

N = final mass of the radioactive sample

No = initial mass of the sample

λ = disintegration constant

t = time for the radioactive decay

In(N/No) = -λt

N / No = e^-λt

N = No(e^-λt)

N = 80 × e^-(0.119 × 20.8)

N = 80 × e^-2.4752

N = 80 × 0.0841

N = 6.728g

The mass of the sample after 20.8 days is approximately 6.73g

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2