Answer:
V CH4(g) = 190.6 L
Explanation:
assuming ideal gas:
∴ STP: T =298 K and P = 1 atm
∴ R = 0.082 atm.L/K.mol
∴ moles (n) = 7.80 mol CH4(g)
∴ Volume CH4(g) = ?
⇒ V = RTn/P
⇒ V CH4(g) = ((0.082 atm.L/K.mol)×(298 K)×(7.80 mol)) / (1 atm)
⇒ V CH4(g) = 190.6 L
Its known as covalently bonded atoms
1 mole of any particles = 6.02* 10²³ particles
4.5*10²⁵ atoms Ni* 1 mol Ni/6.02*10²³ Ni ≈ 74.75≈ 75 mol Ni
