1. How many moles of neon would be contained in 4.0 L of neon gas at 560 kPa and 127oC

The number of moles present is 68267.08 moles

Explanation:

V = 4.0L

P = 560kPa = 560*10³Pa

T = 127°C = (127 + 273.15) K = 400.15K

Number of moles (n) = ?

R = 0.082J / mol. K

From ideal gas equation,

PV = nRT

n = PV / RT

n = (560*10³ * 4.0) / (0.082 * 400.15)

n = 2240000 / 32.8123

n = 68267.08 moles

The number of moles present is 68267.08 moles