Question

Question #560e6

Answer 1

Well, `3xxN_Axx(89.1*g)/(208.23*g*mol^-1)=??`

Explanation:

The quotient `(89.1*g)/(208.23*g*mol^-1)=0.428*mol` gives us the molar quantity....and we know that there are moles of atoms per mole of barium chloride....`2*molxxCl^-` ion, and `1*mol*Ba^(2+)` ion.

And so we take the product...`6.022xx10^23*mol^-1xx0.428*molxx3=7.73xx10^23*"atoms"`

Answer 2

`2.577 * 10^23` molecules of `"BaCl"_2` - or `7.73 * 10^23` atoms.

Explanation:

I had to use Wolfram to find the atomic weight of Barium Chloride.

Wolfram Alpha

This tells me that 1 mole of `"BaCl"_2` is 208.23 grams per mole.

89.1 g is 0.428 of one mole (rounding)

1 mole is `6.023 * 10^23` molecules, which is Avogadro's number. Which is one of the 2 numbers I can remember from my schooling so many years ago.

`0.428 * 6.023 * 10^23 = 2.577 * 10^23` molecules of `"BaCl"_2`.

Each molecule is 1 Barium and 2 chlorine, a total of 3 atoms, so multiply the number of molecules by 3 gives you `7.73 * 10^23` atoms.

GOOD LUCK