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Triple points are typically defined as points on the phase diagram of a pure substance where three phases coexist. Their existence requires that the phase rule be observed, which is $$f=C+2-P$$ where \begin{align}f &: \text{number of degrees of freedom} \\ C &: \text{number of components} \\ P &: \text{number of phases}\end{align} The number ...

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Yes, the triple point of the system will change. Instead of a single point specified by temperature and pressure at which multiple phases of water coexist, the addition of solutes (not solvents) will cause the conditions for 3 phases to exist to become a range of temperatures and pressures.

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It seems to me that it will be difficult to get a uniform concentration of Zn in a non-uniform material like dried leaves, unless you measure only in bulk (this pile of dried leaves weighs X grams and contains 2000X micrograms of Zn). Things get more confusing if the insects don't eat all the dried-leaf structure (leaving stems, for example). I'm also not ...

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In the simplest terms and most convenient definitions, the lowering of vapor pressure is not a colligative property while the relative lowering of vapor pressure is indeed a colligative property. A bit more explanation: A colligative property is dependent on the moles of solute only. The ratio in the definition you provided eliminates the temperature ...

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Water is very unique liquid, because it has a higher density in the liquid state than the solid state. The maximum density of water is found at $\pu{4 ^\circ C}$, which is reported as $\pu{999.9720 kg\:m^{-3}}$ (Temperature Effects on Density). Therefore, molarity of water at $\pu{4 ^\circ C}$ can be calculated as $\frac{\pu{999.9720 g\:L^{-1}}}{\pu{18.015 g\... 1 I've listed a series of points about diffusion and Osmosis. The cause of osmosis is simply diffusion: the solvent can diffuse through the membrane and the solute cannot. Thermodynamically, we say that the reason that the diffusion occurs is that the chemical potential of the solution is at a lower chemical potential than the pure solvent and so there is a ... 2 We can look at solubility as a competition between lattice energy and hydration.$\Delta H_{solution} = \Delta H_{hydration} + \Delta H_{lattice}$If the enthalpy of solution is exothermic or slightly endothermic, the compound will be soluble (we are ignoring entropy here). If we take a simplified view of both lattice energy and enthalpy of hydration, we ... 0 For the lattice energy, both cation and anion radii release more potential energy, when they are smaller. It is similar as if a satellite goes to the lower orbit, or an electron in an atom goes to the quantum state with lower energy. For multiatom anions, it is more complicated due their structure and charge distribution. For hydration energy and cations ... 0 A general idea we have about osmosis is that it is the movement of solvent particles through a semi permeable membrane from a region of high concentration to a region of low concentration. You have to carefully specify which concentration you refer to. The net transport is from regions of high solvent concentration (low solute concentration) to low solvent ... 2 The figure shows the relationship between the solid and liquid solvent and a solution. Instead of freezing at the freezing point of the solvent$T_1$a solution freezes at$T_2$, where the vapour pressure of the solution equals that of the pure solid solvent. Using the Clausius-Clapeyron equation we can show that the solution curve will always intersect ... 2 Vapour pressure is normally defined as an equilibrium phenomenon. In statistical mechanics terms it is the point where an equal number of molecules are leaving the liquid and entering the vapour and leaving the vapour and entering the liquid. This means that, however small the exposed surface, enough molecules have gone into the vapour phase to create the ... 1 This is just one possible approach: First of all, the constants doesn't seem different enough to argue that$[\ce{Ba^{2+}}]=[\ce{S_2O_3}^{2-}]$. So for now consider: $$[\ce{Ba^{2+}}]=[\ce{S_2O_3}^{2-}]+[\ce{SO_4}^{2-}] \hspace{2em} (1)$$ and dividing both$K$s we can find:$\$\frac{K_{\ce{BaS2O3}}}{K_{\ce{BaSO4}}}=\frac{[\ce{S_2O_3}^{2-}]}{[\ce{SO_4}^{2-}]}...

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Answer: Both reactions involving Cu+ have the lowest and highest Standard Electrode Potentials respectively. Thus, Cu+ has to be the strongest oxidizing/reducing agents, and therefore it undergoes disproportionation. Because the formation of Cu has a higher SEP than Cu+, it is more likely to get reduced and therefore is a better oxidizing agent - NOT a ...

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To acknowledge OP's request, I am putting this as my answer: What OP's initially referring to (US20110008445A1) was a patent application. Since it was a 2011 application, I searched for the patent and found it (Ref.1). It's abstract states that: A Suspension of ascorbic acid in glycerol or in glycerol comprising diglycerol, in which the content of ...

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You should proceed as follows: Charge the funnel with the mixture until it is almost full. Let the content settle. Separate the lower phase. Separate the upper phase. Repeat until you have processed all the mixture. The ground joint is to close the funnel hermetically. This is necessary when you need to perform an extraction. A liquid with a mixture ...

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