T₁ = 27°C = 27 + 273 = 300K, V₁ = 6 L,
T₂ = 150°C = 150 + 273 = 423K, V₂ = ?,
By Charles' Law: V₁/T₁= V₂/T₂
6/300 = V₂/423
423*(6/300) = V₂
8.46 = V₂
Volume at 150°C =8.46 L.
1. different gases mix together to form air and no new comound is formed
2. air has variable composition
3. air shows the properties of its constituent substances
Reduction takes place at the cathode and oxidation occurs at the anode.
Anode
B 1.2g/mL just look at a graph and pin point those answers