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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Molar solubility ： ウィキペディア英語版 Molar solubility Molar solubility is the number of moles of a substance (the solute) that can be dissolved per liter of solution before the solution becomes saturated. It can be calculated from a substance's solubility product constant (Ksp) and stoichiometry. The units are mol/L, sometimes written as M. ==Derivation== Given excess of a simple salt AxBy in an aqueous solution of no common ions (A or B already present in the solution), the amount of it which enters solution can be calculated as follows: The Chemical equation for this salt would be: $\_y\}\_\; \backslash Longleftrightarrow\; x\; \backslash text\_\; +\; y\; \backslash text\_\backslash ,$ where A, B are ions and x, y are coefficients... 1. The relationship of the changes in amount (of which mole is a unit), represented as N(∆), between the species is given by Stoichiometry as follows: $-\backslash frac\; =\; \backslash frac\; =\; \backslash frac\backslash ,$ which, when rearranged for ∆A and ∆B yields: $N\_\; =\; -xN\_\backslash ,$ $N\_\; =\; -yN\_\backslash ,$ 2. The Deltas(∆), Initials(i) and Finals(f) relate very simply since, in this case, Molar solubility is defined assuming no common ions are already present in the solution. $N\_\; +\; N\_\; =\; N\_\backslash ,$ The Difference law $N\_\; =\; 0\backslash ,$ $N\_\; =\; 0\backslash ,$ Which condense to the identities $N\_\; =\; N\_\backslash ,$ $N\_\; =\; N\_\backslash ,$ 3. In these variables (with V for volume), the Molar solubility would be written as: $S\_0\; =\; -\backslash frac\backslash ,$ 4. The Solubility Product expression is defined as: $K\_\; =\; ()^x()^y\backslash ,$ These four sets of equations are enough to solve for S0 algebraically: $K\_\; =\; \backslash right)\}^x\; \backslash right)\}^y\backslash ,$ $K\_\; =\; \backslash right)\}^x\; \backslash right)\}^y\backslash ,$ $K\_\; =\; \backslash frac\backslash ,$ $K\_\; =\; \backslash frac\backslash ,$ $K\_\; =\; \backslash frac\backslash ,$ $K\_\; =\; x^x\; y^y\; \backslash frac\backslash ,$ $K\_\; =\; x^x\; y^y\; \backslash frac\}\; =\; \backslash right)\}^\backslash ,$ $\backslash frac\; =\; ^\backslash ,$ Hence; $S\_0\; =\; \backslash sqrt()\}\backslash ,$ 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Molar solubility」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.