The technetium-99 isotope has a half-life of 6.0 hours. If 100.0 mg were injected into a patient how much remains after 18 hours?

12.5mg of technetium-99 remain

Explanation:

When a radioactive isotope decays in the time, its concentration follows the equation:

ln[A] = -kt + ln[A]₀

Where [A] could be taken as the mass of radioactive isotope after time t (18h),

k is rate constant = ln 2 / Half-Life = ln 2 / 6hours = 0.1155hours⁻¹

[A]₀ is initial amount of the isotope = 100.0mg

Replacing:

ln[A] = -0.1155hours⁻¹ *18hours+ ln[100.0mg]

ln[A] = 2.5257

[A] =

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