The half-life of strontium-90 is 28.1 years. estimate the percent of a strontium sample left after 100 years.

Explanation:

1) Half-life is the time for the decay of a sample of matter to half its initial amount.

2) Radioisotopes (radioactive isotopes), such as strontium-90, have constant half-life times.

3) You can relate the time a radioactive samplle has been decaying with the percent of the matter remaining and the number of half-lives elapsed:

Where [A]/[Ao] is the ratio ratio of final amount to initial amoun, and n is the number of half-lives elapsed.

Here, n = 100 years / 28.1 years ≈ 3.559

Then:  

And the percent is 100 × 0.08486 = 84.86% = 84.9 %

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

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