The of nitrogen is 0.060 atm. First, write the balanced chemical equation for the equilibrium. C(s) +2N₂O(g) ⇌ CO₂(g) + 2N₂(g) There was no CO₂ or N₂ present at the...
1 Answers 1 viewsAccording to Dalton's law of partial pressures, the total pressure is equivalent to the sum of the pressures of the individual gases (if they were present alone). Therefore, the...
1 Answers 1 viewsFrom Dalton's Law of , the total pressure of a mixture of gases is the sum of the partial pressures of each gas. Therefore, the partial pressure of $"CO"_2$ is...
1 Answers 1 viewsLet's put it this way: what do you think of assigning oxygen the role of oxidizing agent? Does that make sense? If not, then we can do it another way....
1 Answers 1 views$K_p=(P_(PCl_3)P_(Cl_2))/P_(PCl_5)=0.0121$ Initially, $P_(PCl_5)=0.123*atm$, if a quantity $x$ dissociates, then, $K_P=0.0121=x^2/(0.123-x)$ (I have assumed that the quoted equilibrium constant is $K_P$ not $K_c$. And if I am wrong, I am wrong)....
1 Answers 1 viewsThe equilibrium reaction given to you looks like this $"Xe"_ ((g)) + color(red)(2)"F"_ (2(g)) rightleftharpoons "XeF"_ (4(g))$ You know that the mixture initially contains $"2.24 atm"$ of...
1 Answers 1 viewsDatlon's law of Partial Pressures states that in a gaseous mixture, the exerted by a gaseous component is the same as the pressure it would exert if it ALONE occupied...
1 Answers 1 views$2SO_2+O_2rarr2SO_3$ $K_p=(p_(SO_3)^2)/(p_(SO_2)^2xxp_(O_2))$ $=$ $??$ We need to know $K_p$ in order to access $p_(SO_3)$.
1 Answers 1 viewsI got $"1.6 atm"$. DALTON'S LAW OF PARTIAL PRESSURES The equation that sums up what we have to do in this problem (pun intended) is Dalton's law of partial...
1 Answers 1 viewsJust like we did , this problem just reverses the situation. However, the $DeltaG_f^@$ of $"NO"_2(g)$ is incorrect, and it should be $"51.3 kJ/mol"$, as shown by two sources here:...
1 Answers 1 views