Answer:
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Answer:
(B) SECONDARY ALKYL HALIDE
Answer:
V NH3 = 304.334 L
Explanation:
- 1 mol ≡ 6.02 E23 molecules
⇒ moles NH3 = (7.50 E24 molecules)×(mol/6.022 E23 molecules)
⇒ mole NH3 = 12.454 mol
assuming ideal gas:
STP:
∴ T = 25°C ≅ 298 K
∴ P = 1 atm
⇒ V NH3 = RTn/P
⇒ V NH3 = ((0.082 atm.L/K.mol)×(298 K)×(12.454 mol))/(1 atm)
⇒ V NH3 = 304.334 L