Assuming conditions of standard temperature and pressure, if you have a sample of a noble gas that weighs 12.69 g, in a container with a volume of 14.09 l, which noble gas is it? (r = 0.0821 l∙atm/mol∙k)

Noble gas is Ne

Explanation:

Let's make the expression for the Ideal Gases Law at STP

Temperature = 273K

Pressure = 1 atm

1 atm . 1.409L = n . 0.0821 L∙atm/mol∙K . 273K

1.409 L.atm / (0.0821 L∙atm/mol∙K . 273K) = n

0.0628 mol = n

Let's calculate the molar mass to find out the noble gas

1.269 g / 0.0628 m = 20.2 g/m

BE CAREFUL  BECAUSE THIS ARE WRONG IN THE STATEMENT.

12.69 g

14.09 L

2b2t

explanation:

2b2t

1. a)c-14 has two more protons

2. d)they have different numbers of protons, neutrons, and valence electrons; they are all gases at room temperature.

3. c) iron and nickel.