I know the answer but don't know how to work it... here goes...
Gaseous argon has a density of 1.40 g/L at standard conditions. How many argon atoms are in 1.00 L of argon at standard conditions?
Should the answer be 2.1*10 to the 22nd power?
2006-10-30 13:10:55 · 3 answers · asked by ink_spell_020 1 in Science & Mathematics ➔ Chemistry
Checks out when you divide 1.40 by weight of argon and multiply by Avogadro's number.
2006-10-30 13:18:26 · answer #1 · answered by Anonymous · 0⤊ 0⤋
the mass of 1.00L argon 1.40 * 1.00 = 1.40 g
the moles of argon 1.40 / 40 = 0.035 moles
the amount of argon atoms : 0.035 * 6.02 * 10^23 = 2.107 *10^22
2006-10-30 21:19:17 · answer #2 · answered by James Chan 4 · 0⤊ 0⤋
a mole of any gas occupies 22.4 liters. So if you have 1 liter you have 1 / 22.4 moles. 6.02x 10 to the 23rd atoms per mole.
NOW MULTIPLY
2006-10-30 21:27:19 · answer #3 · answered by science teacher 7 · 0⤊ 0⤋