How many moles are in 79.85g fe2o3?

in 79.85 g of Fe2O3 there will be 0.5 moles of Fe2O3

Explanation:

To find the number of moles we must calculate the molecular weight of iron oxide, for that we use the periodic table:

mFe = 55.845 g/mol

mO = 15.999 g/mol

mFe2O3 = (2x55.845) + (3x15.999) = 159.687 g/mol

1 mole of iron oxide weighs 159.687 g, therefore, to calculate the number of moles present in 79.85 g it will be equal to:

nFe2O3 = 79.85/159.687 = 0.5 moles

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)