V = 8963 L
Our strategy here is to realize this problem involves stoichiometric calculation based on the balanced chemical equation for the combustion of octane:
C₈H₁₈ + 25/2 O₂ ⇒ 8 CO₂ + 9 H₂O
From the density of octane we can obtain the number of moles:
D = m/V ⇒ m= D x V = 0.7025 g/mL x ( 1000 mL) = 702.5 g
MW octane = 702.5 g/ 114.23 g/mol = 6.15 mol
Required mol oxygen to react with octane:
6.15 mol octane x 25/2 mol O₂ / mol octane = 76.8 mol O₂
Now mol fraction is given by mol O₂ / total number mol air ⇒
n air = 76.8 mol O₂ / (0.2095 mol O₂ / mol air ) = 366.59 mol air
and from the ideal gas law we can compute the volume of air:
PV = nRT ⇒ V = nRT/P
V = 366.59 mol air x 0.08205 Latm/Kmol x (25+ 273) K/ 1 atm
= 8,963 Lts
Note we treat here air as a compund which is allowed in combustion problems.
Potassium nitrate changes the most to from 0 C to 100 C
Adhesive forces and cohesive forces
Cohesive forces is how much the water molecules stick to each other and adhesive forces are to which extinct they stick to other surfaces.
When we pull the paperclip upward the water molecules adhere to it due to adhesive forces. While the water on the clip is remain connected to water in the glass because of cohesive forces.
So if the adhesive forces dominates the water will be pulled out and if the cohesive forces dominate it will remain with other water in the cup.
The order of reaction is 2.
Rate constant is 0.0328 (M s)⁻¹
The rate of a reaction is inversely proportional to the time taken for the reaction.
As we are decreasing the concentration of the reactant the half life is increasing.
a) For zero order reaction: the half life is directly proportional to initial concentration of reactant
b) for first order reaction: the half life is independent of the initial concentration.
c) higher order reaction: The relation between half life and rate of reaction is:
Half life =
where n = order of reaction
Hence n = 2
K = 0.0328