correct me if I'm wrong but I think the component of the applied force along the inclined surface should be ##Mgsinθ## as well (just to the opposite direction).
I spoke to my Chemistry teacher about it and she said that since NaOH is already VERY soluble in water the increase in temperature shouldn't have a noticeable effect on its equilibrium.
does it matter that solubility increases if in both cases all of the NaOH dissolves?
edit: or well not ALL of it but in both cases, it reaches the equilibrium
ok so let me know if I understand right (from what I read in the link you shared).
In exothermic reactions - as the initial temperature increases, dH/R becomes smaller hence dH being a smaller value?
so if I apply this to the example I gave with NaOH - as I increase the initial temperature of...
Say I dissolve NaOH in water - a very exothermic reaction.
Does it matter if I dissolve it in water that is at room temperature or at a different temperature?
Will changing the initial temperature of the water affect the dH of the reaction (hence affect the change in temperature at the end of...