How many grams of argon gas would be in a 0.67 L container at a 0.50 atm and a 7.8C° (R=0.0821)

the answer is the first option

explanation:

just googled it m8. not that hard

lol

increased stability in the completion of fluorine's outer shell.

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer