Heat produced = -13588.956 kJ
<h3>Further explanation</h3>
Given
The reaction of combustion of Methane
CH4(g)+2O2(g)→CO2(g)+2H2O(g) ΔH∘rxn=−802.3kJ
271 g of CH4
Required
Heat produced
Solution
mol of 271 g CH₄ (MW=16 g/mol0
mol = mass : MW
mol = 271 : 16
mol = 16.9375
So Heat produced :
= mol x ΔH°rxn
= 16.9375 mol x −802.3kJ/mol = -13588.956 kJ
To determine the upper bond
Ec(u) = EmVm + EpVp
Em is the elastic modulus of cobalt.
E₁ is the elastic modulus of the particulate
Vm is the volume fraction of cobalt
Vp is the volume fraction of particulate
substitute
Ec(u) = 200 (Vm) + 700 (Vp)
To determine the lower bound
Ec (l) = EmEp/VmEp+ VpEm
Substitute
Ec (l) = 200(700)/Vm(700) + Vp (200)
Ec (l) = 1400/7Vm+2Vp
I cant help unless there is an article..
Answer:
i believe it is strong acids, strong bases, and salts.
Explanation:
Hope this helps : )
