1.00 g of he, 14.0g f2, and 19.0 g ar are placed in a 13.0-l container at 20.0 °c. the total pressure (in atm) in the container is atm.

The total pressure in the container is 2.02 atm.

Step 1. Calculate the total number of moles.

Moles of He = 1.00 g He × (1.00 mol He/4.003 g He) = 0.2498 mol He

Moles of F₂ = 14.0 g F₂ × (1 mol F₂/38.00 g F₂) = 0.3684 mol F₂

Moles of Ar = 19.00 g Ar × (1 mol Ar/39.95 g Ar) = 0.4756 mol Ar

Total moles = (0.2498 + 0.3684 + 0.4756) mol = 1.094 mol

Step 2. Calculate the total pressure.

Use the Ideal Gas Law: pV = nRT

We can solve the equation to get the pressure:

p = (nRT)/V

n = 1.094 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = (20.0 + 273.15) K = 293.15 K

V = 13.0 L

∴ p = (1.094 mol × 0.082 06 L·atm·K⁻¹mol⁻¹ × 293.15 K)/13.0 L = 2.02 atm

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2

it is a solid

explanation: