An iron ore was found to contain 92.0% pyrite (fool's gold). pyrite has the formula fes2and is 46.5% iron. how many grams of iron are there in 52.5 grams of iron ore? give only numerical value in the box below (no units)

There are 22.46 g of iron in 52.5 g of iron ore.

Explanation:

Iron ore:

92.0 FeS₂    100 g iron ore

      x            52.5 g iron ore

               x = 48.3 g FeS₂

Pyrite:

100 g FeS₂   46.5 g Fe

48.3 g FeS₂     y

              y = 22.46 g Fe

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally

c. 4.54 x 10–12 j happens to be correct