Answer:
5.94
Explanation:
At equilibrium
⇒[CH3NH^+] = 0.130/2= 0.065 M
hence we can write that
x= concentration of H+ ions
⇒2×10^(-11)×0.065 = x^2
⇒x= 1.14×10^{-6}
therefore ph=log [H+]= log[ 1.14×10^{-6}]= 5.94
Answer:
Would be increasing the concentration of SO3 as you chose in the question.
Explanation:
Increasing the concentration of the reactant will most likely shift it to the right by adding more products.
Answer:
- 58 kJ
Explanation:
The equation of the reaction is given as;
2NO2(g) → N2O4(g)
N2(g) + 202(g) → 2NO2(g) delta H = 67.7 kJ
Since NO2 is the reactant in the equation, we have to reverse this reaction. We now have;
2NO2(g) → N2(g) + 202(g) delta H = - 67.7 kJ
N2(g) + 202(g) → N2O4(g) delta H = 9.7 kJ
Since NO4 is the product in the equation, we don't have to do anything to this reaction.
Adding the two reactions;
2NO2(g) → N2(g) + 202(g) delta H = - 67.7 kJ
N2(g) + 202(g) → N2O4(g) delta H = 9.7 kJ
---------------------------------------------------------------- N2(g) and 2O2 (g) cancels out
2NO2(g) → N2O4(g) delta H = (-67.7 + 9.7) = - 58 kJ
In a model, the particles are not the actual size. They are just representing the gas particles. In a gas, the particles are smaller and in their own material