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2

The Lattice and hydration energies applies to cases of dissolution of ionic compounds like $\ce{NaCl}$. For covalent liquid compounds with polar bonds like water, the lattice energy is replaced by the bond dissociation energy. The value of the dissociation energy ( enthalpy ) -- involving hydration -- can be indirectly determined from the temperature ...

0

The amount of your substance is $\pu{0.16 mol}$. The mass of your substance is $10\ \mathrm g$. So the molar mass is: $M=10\ \mathrm g/0.16\ \mathrm{mol}=62.6\ \mathrm{g/mol}$. With such a molar mass, your substance may be nitric acid $\ce{HNO3}$, whose molar mass is $\pu{63 g/mol}$. But this choice is improbable, because, in aqueous solution, nitric acid is ...

0

Your initial solution consists of 30 kg NaOH in 1000 kg of water; it will treat a certain amount of bark ribbon. I suspect that the final pH of the solution is much less than the original pH; i.e., the solution gets used up by reacting with the plant material. You mention that one of the biggest costs is for NaOH. Optimizing the procedure could involve the ...

4

The equilibrium is independent on the pressure above the liquid because the mechanism of creating it has nothing to do with the other gases above the liquid It may seem intuitive that a higher gas pressure above a liquid would "push" liquid vapour back into the liquid. But not if you understand how gases work or how a liquid/vapour equilibrium is ...

3

I assume you consider the saturated vapor pressure at 100% relative humidity. (As at 50% rel.humidity, vapor pressure = 0.5 saturated vapor pressure). Be aware that ability to contain certain amount of vapor, having respective vapor pressure, is property of space, not of a gas. In first approximation, presence of air or other gas has no effect of ...

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