Question #560e6

2 Answers
Dec 7, 2017

Answer:

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"#

Dec 7, 2017

Answer:

#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