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Sodium trioxocarbonate (iv) pentahydrate, a.k.a washing soda, is an efflorescent crystal containing 10 molecules of water of crystallization in its molecular formula.The molar mass of washing soda comprises a: carbonate portion of 106 g + pentahydrate portion of 180 g = 286 g/mol. i.e. 1 mole or 286 g of washing soda contains 106 g of ...


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The molar mass of copper (ii) sulfate penta-hydrate is $\pu{249.69 g/mol}$ and that for anhydrous copper(ii) sulfate is $\pu{159.61 g/mol}$. Thus, in $1$ mole of the hydrated substance there are $\pu{159.61 g}$ of copper (ii) sulfate and $5 \times \pu{18.00 g}$ (i.e. $\pu{90.00 g}$) of molecules of water of crystallization. $\pu{5M}$ copper (ii) sulfate ...


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$\ce{P2O5}$ content is useful as it is the phosphorous content that a scientist often cares about when using phosphoric acids. This might seem silly given that as you pointed out anything from $0-70\% \ \ce{P2O5}$ could just be expressed as a percentage of $\ce{H3PO4}$ but there are times when a higher $\ce{P2O5}$ concentration is desired. For example ...


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Your question is a valid analytical chemistry question which has deep roots in history. No, the reason is not that $\ce{P2O5}$ reacts with water to from phosphoric acid and that is why $\ce{P2O5}$ values are quoted as suggested in the comments below the question. This is a very old historical tradition. In older analytical chemistry, chemists had only two ...


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Generally, it is preferable to write down the complete quantity equation before plugging in the numbers. If necessary, perform all mathematic operations on quantities symbolically. Do not plug in the numbers until you have only one equation for the desired result. Always write the values with the correct units and carry the units through the calculation. Do ...


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$\ce{OH-}$ can't be reduced. It may only be oxidized by giving an electron at the anode, while $\ce{H+}$ is reduced instead of $\ce{OH-}$. Secondly, the reducing property of $\ce{H2O}$ is greater than that of $\ce{H+}$ ion. Moreover, it is greater in number than other ions like $\ce{H+}$ ions etc. So it will be preferentially discharged although it is ...


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As Ed said you are following the correct protocol by diluting the sample. The key reason is for doing so is to avoid a non-linear region in the calibration curve. This is why one would dilute the sample rather than make more concentrated standards. Several interesting effects happen with alkali metals at high concentrations in the flame or plasma. One of ...


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A solution by definition is a homogeneous mixture, so its "concentration" is same throughout the bulk. Let us have a look as to how one would prepare a solution of a given concentration, say in 1 L. We would weigh out a known weight of a solute and dissolve it ~ 500 mL of solvent, with mechanical stirring or shaking. We would add more solvent until the ...


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