The graph below shows the half life values of parent isotopes. The bar graph shows 'Parent Isotope' labeled on the x axis and 'Half-life (in billion years)' labeled on the y axis. The first bar is labeled Uranium-238 and its height is 4. The second bar is labeled 'Thorium-235' and its height is 14, the third bar is labeled 'Potassium-40' and its height is 1.5', and the fourth bar is labeled 'Rubidium-87' and its height is 49.

Based on the graph, it can be concluded that the isotope which is most likely to be found with 6.25% of its original amount in 6 billion years is

Uranium-238
Thorium-235
Potassium-40
Rubidium-87

Looking up graphs on google, you can see thaty = force(n)x = time(s)

Unbalanced equation: fe2o3+ cofe + co21 fe2o3 + 3 co = 2 fe(s) + 3 co212.8g fe2o3 x (1 mol fe2o3 / 159.7g fe2o3) = 0.08015 mol fe2o311.5g co x (1 mol co / 28.01g co) = 0.4106 mol cofrom the balanced equation, 1 fe2o3 reacts with 3 cosince we have 0.08015 mol fe2o3 we therefore need0.08015 mol fe2o3 x (3 mol co / 1 mol fe2o3) = 0.2404 mol cosince we have 0.4106 mol co but only need 0.2404 to completely consume all thefe2o3, we have xs (excess) co and fe2o3 is therefore the limiting reagentfrom the balanced equation, 1 fe2o3 = 2 fe(s)0.08015 mol fe x (2 mol fe(s) / 1 mol fe2o3) = 0.1603 mol fe(s)theoretical mass of fe .1603 mol fe x (55.84g fe / mole fe) = 8.951g fe% yield = (6.10g / 8.951g) x 100% = 68.1%i didn't do this work, so i am not one hundred percent sure if it is correct  seems like a reliable website,