ΔH(reaction) = ΔH(formation of products) - ΔH(formation of reactants)
ΔH(reaction) = ( 1*ΔH(Pb(s)) + 1*ΔH(CO2(g)) ) - ( 1*ΔH(PbO(s)) + 1*ΔH(CO(g)) )
ΔH(reaction) = ( 0 + -393.5 ) - ( ΔH(PbO(s)) + -110.5 )
ΔH(reaction) = -283 - ΔH(PbO(s))
-131.4 = -283 -ΔH(PbO(s))
ΔH(PbO(s)) = -151.6 kJ
So, the best answer is A.
Answer:
Explanation:
Here we have the mass of CO₂ added = 340 g
From
We have, where the molar mass of CO₂ is 44.01 g/mol
Therefore,
71. Included drawing attached
72. Here we have the pressure of the gas given by Charles law which can be resented as follows;
Where:
P₁ = Initial pressure = 6.1 atmospheres
P₂ = Final pressure
T₁ = Initial Temperature = 293 K
T₂ = Initial Temperature = 313 K
Therefore,
The combustion of any hydrocarbon yields water and carbon dioxide. We will now construct a balanced equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Each mole of propane requires 5 moles of oxygen.
Answer:
d , before the molten rock becomes lava, it is first magma, and most people know that lava is ejected from volcanoes
Answer:
0.1 M
Explanation:
The overall balanced reaction equation for the process is;
IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)
Generally, we must note that;
1 mol of IO3^- require 6 moles of S2O3^2-
Thus;
n (iodate) = n(thiosulfate)/6
C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6
Concentration of iodate C(iodate)= 0.0100 M
Volume of iodate= V(iodate)= 26.34 ml
Concentration of thiosulphate= C(thiosulfate)= the unknown
Volume of thiosulphate=V(thiosulfate)= 15.51 ml
Hence;
C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)
0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M
