An iron bar weighed 565 g. After the bar had been standing in moist air for a month, exactly one-eighth of the iron turned to rust (Fe2O3). Calculate the final mass of the iron bar and rust.

Final mass of Iron bar will be 621.105 g.

Mass of rust = 126.730 g

Explanation:

Exactly one- eighth of the Iron turns to rust. That means, the mass of Iron  that under go the process of rusting() = 565 × = 70.625 g

So, the mass of pure Iron after rusting will be = 565 - 70.625 = 494.375 g

Now the reaction of rusting as follows,

   2Fe + →

Now, molecular weight of Iron = 55.84 g

So, moles of Iron undergoes the reaction is = = 1.26 mol

Molecular weight of = 159.68 g

Here 2 moles of Iron undergoes the reaction to produce 2 moles of Iron oxide.

Hence, 1.26 mol of Iron will produce 1.26 mole of Iron oxide.

Mass of = = 126.730 g

∴Mass of rust = 126.730 g

∴ Final mass of Iron bar = ( 494.375 + 126.730) = 621.105 g

hope this

explanation:

covalent bonds

consist of pairs of electrons shared by two atoms, and bind the atoms to fixed orientation and requre high energies to break them

ionic bonds

when two atoms differ considerably in their electronegativity, then one of the atoms will lose its electron to the other atom. this results in a positively charged ion(cation) and a negatively charged ion (anion)

it's c. elements are placed in rows based on how may valencies electrons have.