Nitrous oxide (n2o), or laughing gas, is commonly used as an anesthetic in dentistry and surgery. how many moles are present in 8.4 liters of nitrous oxide at stp?

Answer is: 0.375 moles are present in 8.4 liters of nitrous oxide at stp.

V(N₂O) = 8.4 L.

V(N₂O) = n(N₂O) · Vm.

Vm = 22,4 L/mol.
n(N₂O) = V(N₂O) ÷ Vm.

n(N₂O) = 8.4 L ÷ 22.4 L/mol.

n(N₂O) = 0.375 mol.
Vm - molare volume on STP.

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

no but i do not usually comment or use my voice that much when it comes to social media