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1

You have replicated electro-rafination of copper. Copper ions from the being dissolved anode are replacing the copper ions being deposited on the cathode. Therefore overall electroneutrality of the solution is kept. The local neutrality is managed by electromigration of both ions $\ce{Cu^2+}$ and $\ce{SO4^2+}$.


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Trial and error, but most often the boiling point of the lower-boiling solvent. The absolutely correct answer would be to look at a phase diagram, determine whether you have a positive or negative azeotope (or non at all), find out where on the phase diagram you are and then derive the temperature. That would tell you approximately at what temperature your ...


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The description you provided is essentially a shorthand for a standard workup procedure. After performing a reaction, one wants to isolate one’s product while removing the catalyst or other reagents and side products that one doesn’t need. For example, in many cross-coupling reactions an inorganic halide salt is an undesired side-product and a base such as ...


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The solvent of this experiment should be non polar, and may be a hydrocarbon, or a chlorinated hydrocarbon. This organic phase may contain as impurities some acids and other polar substances soluble in water, that should be eliminated. This liquid is then mixed with an aqueous solution of soda $\ce{NaHCO3}$. Aqueous solutions and non polar substances are not ...


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This is standard for purifying substances. To wash means to add your product solution to an aqueous solution (or just water, but frequently a saturated solution) to a separatory funnel. After shaking, you drain the lower layer (which is usually aqueous). This process removes water soluble impurities. This is frequently repeated. Drying is accomplished by ...


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I'd like to specifically commment on this: According to the NCERT for Class XII, Part I, [pg. 46, para 3][1], In a pure liquid the entire surface is occupied by the molecules of the liquid. If a non-volatile solute is added to a solvent to give a solution [Fig. 2.4.(b)], the vapor pressure of the solution is solely from the solvent ...


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Vapor pressure is the pressure exerted by the vapors when in dynamic equilibrium with its condensed phase at a given temperature. It is the maximum value of partial pressure at a given temperature. So if pressure is increased after this value then some vapors will condense and if pressure is decreased then more liquid will vaporise until the vapor pressure ...


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tl;dr It's so simple it was likely never published as a "research result", since it stems quite directly from definitions. It's a small step from the conservation of matter and the definition of concentration. By definition: $C = \frac{n}{V}$, thus $n = CV$. If you dilute with pure solvent, you don't change the amount of solute that you have, in ...


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For your future assignment it is always worth work with two half-reactions: oxidation and reduction: Given is: $$\ce{NiO2(s) + 4 H+(aq) + 2 Ag(s) -> Ni^2+(aq) + 2 H2O(l) + 2 Ag^+(aq)}\quad E^\circ = \pu{2.48 V}$$ Oxidation half-reaction: $$\ce{Ag (s) <=> Ag+ + e-} \tag1$$ Reduction half-reaction: $$\ce{NiO2 (s) + 4 H+ + 2e- <=> Ni^2+ + 2H2O (l)...


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Colloidal solutions scatter light of different wavelenghts in different directions, and that's also dependent on the size(!) of the colloidal particles. To see any scattering at all, there only has to be a difference in the refractive index of pure solution and dissolved particles. The individual colour (i.e. absorption spectrum) of the particles is ...


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Certain solutes,especially under pressure, and actually become incorporated into the water-ice structure to form clathrate hydrates. Of these, methane hydrate is the best known, occurring naturally in oceans where organisms generate the methane under sufficient pressure to create the clathrate phase. These illustrations from the source linked above shows ...


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The $\mathrm{pH}$ of your final ammonia solution is $11.0$. When you dissolve ammonia in water it tends to ionize with water to give following equilibrium: $$\ce{NH3 + H2O <=> NH4+ + OH-} \quad K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} = 1.8 \times 10^{-5}$$ Suppose final concentration of $\ce{NH3}$ is $c$, and equlibrium concentrations ...


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