Answer:
The answer is 102.3!
Explanation:
you get this by multiplying 34.1 x 3 to get 102.3.
Wondering why you multiply 34.1 times 3?
WELLLLLLLLLLL...
when you get 1 mole of h202, you get 34.1 so if you ask for 3 moles of H202, you get 102.3!
<span>Answers are:
-4 for C in CH4, because carbon has greater electronegativity than hydrogen and he attracts shared electrons.
</span><span>+4 for C in CO2, because carbon has smaller electronegativity than oxygen.
</span><span>+1 for H in both CH4 and H2O, because hydrogen has amaller electronegativity than both carbon and oxygen.
</span>
Solid and liquid particles
Answer:
A. ![K=\frac{[N_2O]^2}{[N_2]^2[O_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BN_2O%5D%5E2%7D%7B%5BN_2%5D%5E2%5BO_2%5D%7D)
Explanation:
Hello there!
In this case, for us to figure out the appropriate equilibrium expression, it will be firstly necessary for us to recall the law of conservation of mass which states that the equilibrium constant of an equilibrium chemical reaction is written by dividing the products and reactants and including the stoichiometric coefficients as exponents. In such a way, for the given reaction, we will have:
As N2O is the product whereas N2 and O2 are reactants; thus, the equilibrium expression will be A.
Regards!
Answer:
the standard enthalpy of formation of this isomer of C₈H₁₈ (g) = -375 kj/mol
Explanation:
The given combustion reaction
C₈H₁₈ + 25/2 O₂ → 8 CO₂ + 9 H₂O ............................(1)
Heat of reaction or enthalpy of combustion = -5099.5 kj/mol
from equation (1)
ΔH⁰reaction = (Enthalpy of formation of products - Enthalpy of formation of reactants)
Or, - 5099.5 = [8 x ΔH⁰f(CO₂) +9 x ΔH⁰f(H₂O)] - [ΔH⁰f(C₈H₁₈) + ΔH⁰f(O₂)].................................(2)
Given ΔH⁰f(CO₂) = - 393.5 kj/mol & ΔH⁰f(H₂O) = - 285.8 kj/mol and ΔH⁰f(O₂)= 0
Using equation (2)
ΔH⁰f(C₈H₁₈) = -621 kj/mol
