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The reaction of interest is: $$\ce{XY(s) <=> X(g) + Y(g)} \tag1$$ Thus, $K_p = P_\ce{X}\cdot P_\ce{Y} = 4.1$ since $P_\ce{XY} = 1$ in given condition because it is a solid. If we assume $\ce{X}$ and $\ce{X}$ are real gases, the given conditions are such that $P_\ce{X} = \frac{n_\ce{X} RT}{V} = n_\ce{X}$ and $P_\ce{Y} = \frac{n_\ce{Y} RT}{V} = n_\ce{Y}$,...


3

It is important to know the externally caused change the system is dealing with. If the primary change is increasing the pressure, then the equilibrium shifts to the lower volume, in case of gases the lower molar amount. If the primary change is increasing the temperature, than the equilibrium shifts away from the direction that evolves thermal energy. If ...


2

The electrode redox reactions of the Daniell cell: $$\ce{Zn(s) <=> Zn^2+ + 2 e- }$$ $$\ce{Cu(s) <=> Cu^2+ + 2 e- }$$ maintain at the respective electrode a particular potential where both opposite reactions have the same rate, implying there are no external galvanic causes that affect this potential. Rates of oxidation electrode reactions ...


2

BTW, electrons don't actually actually "flow" very fast. The drift velocity of electrons is incredibly slow, on the order of a millimeter every 5 or ten seconds in a room-temperature conductor. The propagation of current is like the particles in Newton's Cradle, where the electric field bumps each electron down the line, a much faster process. ...


2

Following a discussion that took place in the comments: If given answers are correct, as Poutnik pointed out, your equation should be : $$\require{cancel}\ce{A(g) + B(g)<=>2 C(g)}$$ - Mathew Mahindratne Could you explain how you arrived at that conclusion? - Ollie This answer solves this question for a general case: $$\ce{a A(g) + b B(g) <=> c ...


2

Since you are given the volume, temperature and initial amount of each gas, you can compute initial partial pressures as $n_iRT/V$ and from these the product $Q_p=p_X\cdot p_Y$ and compare this to $K_p$. If $Q_p>K_p$ then solid will form: a. $\pu{5.0 mol}$ of $\ce{X}$, $\pu{0.5 mol}$ of $\ce{Y}$ $Q_p=2.50 \rightarrow$ no solid is formed b. $\pu{2.0 mol}$ ...


1

The molar enthalpy of formation at room temperature is not dependent on the reaction effectively happening at the room temperature. It follows the general principle of state variables, being related to the Hess's law, which refers to the fundamental law of energy conservation. The principle says, that the change of enthalpy ( or any other state variable ) ...


1

In an x-y diagram, the "bulging out" is demonstrating the increase in relative volatility of the components; the number of stages is a strong function of relative volatility, so an increase in relative volatility leads to less stages.


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What happens in salt hydrolysis is enhancement of water ionization with salts. Try asking yourself where do $\ce{H+}$ and $\ce{OH-}$ come from in the solution of $\ce{AB}$, and you'll find that they come from $\ce{H2O}$. Or equivalently, hydrolysis of $\ce{A-}$ can be written as $$ \ce{A- + H+ <=> HA,H2O <=> H+ + OH-}\stackrel{\text{add}}{\...


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As already said in the comments, you cannot use any unit you want in the reaction quotient; it has to be formulated in terms of activity (for solutes) or fugacity (for gases). The "standard state" of a chemical species is a reference state in which the properties of the chemical species are defined. This standard state is a theoretical state, it ...


1

When the pH is equal to the pKa, you have high concentrations of both conjugate acid and conjugate base. You need the weak base to react with added acid, and the weak acid to react with added base to stabilize the pH. When the pH is very different from the pKa, one of those two species will be at very very low concentration (there is an exponential ...


1

Why does a buffer work best at the $\mathrm{pH}$ closest to its $\mathrm{p}K_\mathrm{a}$? For the novices like you who just begin to learn chemistry, this question can be easily explain by using Henderson Hasselbalch equation. Suppose you have an aqueous weak acid solution ($\ce{HA}$). It will be in ionic equilibrium as follows: $$\ce{HA (aq) <=> H+ (...


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