Alpha Caeli - Revision history

98.230.200.132: :Category:Pages using duplicate arguments in template calls

2014-12-24T03:21:55Z

Category:Pages using duplicate arguments in template calls

← Older revision		Revision as of 05:21, 24 December 2014
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	~~{{Unreferenced\|date=December 2009}}~~		Let me initial start by introducing myself. My title is Boyd Butts even though it is not the title on my beginning [http://rhrealitycheck.org/article/2013/02/22/trich-the-most-common-sexually-transmitted-infection-you-may-never-have-heard-of/ certification]. over the [http://Depts.washington.edu/handbook/syndromesBoth/ch9_ulcers.html counter std] test ([http://www.gaysphere.net/blog/262619 More suggestions]) favorite hobby for std testing [http://www.streaming.iwarrior.net/user/WMcneil at home std testing] [http://www.youporntime.com/blog/12800 home std test] my kids and me is to perform baseball and I'm attempting to make it a profession. Years ago we moved to North Dakota over the counter std test and I adore each day living here. For many years he's been operating as a meter reader and it's something he really enjoy.
	~~'''Molar [[solubility]]'''~~ is the ~~number of~~ [~~[Mole (unit)\|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 (K<sub>sp<~~/~~sub>) and [[stoichiometry]]~~. ~~The units are mol~~/~~L, sometimes written as [[Molar (concentration)\|M]].~~

	~~==Derivation==~~
	~~Given excess of a simple salt A<sub>x<~~/~~sub>B<sub>y<~~/~~sub> 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:<br>~~
	~~<math>{\text{A}_x \text{B}_y}_{(s)} \Longleftrightarrow x \text{A}_{(aq)} + y \text{B}_{(aq)}\,<~~/~~math><br>~~
	~~where A, B are ions and x, y are coefficients...<br>~~
	~~<br>~~
	~~1. The relationship of the changes in amount (of which [[Mole (unit)\|mole]] is a unit), represented as N<sub>(∆)<~~/~~sub>, between~~ the ~~species is given by [[Stoichiometry]] as follows:<br>~~
	~~<math>~~-~~\frac{N_{ AxBy(\Delta)}}{1} = \frac{N_{A(\Delta)}}{x} = \frac{N_{B(\Delta)}}{y}\,</math><br>~~
	~~which, when rearranged for ∆A and ∆B yields:<br>~~
	~~<math>N_{A(\Delta)} =~~ -~~xN_{AxBy(\Delta)}\,</math><br>~~
	~~<math>N_{B(\Delta)} =~~ -~~yN_{AxBy(\Delta)}\,<~~/~~math><br>~~
	~~<br>~~

	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.<br>~~
	~~<math>N_{(i)} + N_{(\Delta)} = N_{(f)}\,</math>~~ [~~[Difference operator\|The Difference law]]~~

	~~<math>N_{A(i)} = 0\,<~~/~~math><br>~~
	~~<math>N_{B(i)} = 0\,<~~/~~math>~~

	~~Which condense to the identities<br>~~
	~~<math>N_{A(f)} = N_{A(\Delta)}\,<~~/~~math><br>~~
	~~<math>N_{B(f)} = N_{B(\Delta)}\,<~~/~~math><br>~~
	~~<br>~~

	~~3. In these variables (with V for volume), the Molar solubility would be written as:<br>~~
	~~<math>S_0 = -\frac{N_{ AxBy(\Delta)}}{V}\,<~~/~~math><br>~~
	~~<br>~~
	4. ~~The [[Solubility Product~~]~~] expression is defined as:<br>~~
	~~<math>K_{sp} = [A]^x[B]^y\,</math><br>~~
	~~<br>~~


	~~These four sets of equations are enough to solve for S<sub>0</sub> [~~[~~algebra]]ically~~:~~<br>~~
	~~<br>~~
	~~<math>K_{sp} = {\left(\frac{N_{A(f)}}{V}\right)}^x {\left(\frac{N_{B(f)}}{V}\right)}^y\,<~~/~~math>~~

	~~<math>K_{sp} = {\left(\frac{N_{A(\Delta)}}{V}\right)}^x {\left(\frac{N_{B(\Delta)}}{V}\right)}^y\,<~~/~~math>~~

	~~<math>K_{sp} = \frac{{(N_{A(\Delta)})}^x {(N_{B(\Delta)})}^y}{V^{(x+y)}}\,</math>~~

	~~<math>K_{sp} = \frac{{(-xN_{AxBy(\Delta)})}^x {(-yN_{AxBy(\Delta)})}^y}{V^{(x+y)}}\,<~~/~~math>~~

	~~<math>K_{sp} = \frac{{(-1)}^x {(x)}^x {(N_{AxBy(\Delta)})}^x {(-1)}^y {(y)}^y {(N_{AxBy(\Delta)})}^y}{V^{(x+y)}}\,<~~/~~math>~~

	~~<math>K_{sp} = x^x y^y \frac{{(-1~~)~~}^x {(-1)}^y {(N_{AxBy(\Delta)})}^x {(N_{AxBy(\Delta)})}^y}{V^{(x+y)}}\,<~~/~~math>~~

	~~<math>K_{sp} = x^x y^y \frac{{(-1)}^{(x+y)}{(N_{AxBy(\Delta)})}^{(x+y)}}{V^{(x+y)}}\,<~~/~~math>~~

	~~<math>\frac{K_{sp}}{x^x y^y} = {\left(-\frac{N_{AxBy(\Delta)}}{V}\right)}^{(x+y)}\,<~~/~~math>~~

	~~<math>\frac{K_{sp}}{x^x y^y} = {\left(S_0\right)}^{(x+y)}\,<~~/~~math>~~

	~~Hence;~~

	~~<math>S_0 = \sqrt~~[~~(x+y)]{\frac{K_{sp}}{x^x y^y}}\,<~~/~~math>~~

	~~==Simple calculation==~~
	~~If the [[solubility product]] constant (K<sub>sp<~~/~~sub>) and dissociation product ions are known, the molar solubility can be computed without the aforementioned equation~~.

	~~===Example===~~
	~~Ionic substance AB<sub>2</sub> dissociates into A and 2B, or one mole of ion A and two moles of ion B~~. ~~The soluble ion dissociation equation can thus be written as:~~

	~~AB<sub>2<~~/~~sub> --> A + 2B~~

	~~where the corresponding solubility product equation is:~~

	~~K<sub>sp<~~/~~sub> = [A~~]~~[B]^2~~

	~~If the initial concentration of A~~ is ~~x, then that of B must be 2x~~. ~~Inserting these initial concentrations into~~ the ~~solubility product equation gives~~

	~~K<sub>sp</sub> = (x)(2x)^2~~

	~~If Ksp is known, x can be computed, which is the molar solubility~~.

	~~{{DEFAULTSORT:Molar Solubility}}~~
	~~[[Category:Solutions]]~~

en>Trappist the monk: Fix CS1 deprecated date parameter errors (test) using AWB

2014-02-02T18:28:17Z

Fix CS1 deprecated date parameter errors (test) using AWB

← Older revision		Revision as of 20:28, 2 February 2014
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	~~Nice to satisfy you~~, ~~I am Marvella Shryock~~. ~~Supervising is my occupation~~. ~~One~~ of the ~~issues she enjoys most~~ is ~~to study comics~~ and ~~she'll be starting something else alongside with it~~. ~~Puerto Rico~~ is ~~exactly where he's always been residing but she requirements~~ to ~~transfer because of her family~~.<br><br>~~Here~~ is ~~my web~~-~~site at home std testing~~ ([~~http:~~//~~Www~~.~~Revleft~~.~~com~~/vb/~~member~~.~~php?u~~=~~160656 More~~])		{{Unreferenced\|date=December 2009}}
			'''Molar [[solubility]]''' is the number of [[Mole (unit)\|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 (K<sub>sp</sub>) and [[stoichiometry]]. The units are mol/L, sometimes written as [[Molar (concentration)\|M]].

			==Derivation==
			Given excess of a simple salt A<sub>x</sub>B<sub>y</sub> 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:<br>
			<math>{\text{A}_x \text{B}_y}_{(s)} \Longleftrightarrow x \text{A}_{(aq)} + y \text{B}_{(aq)}\,</math><br>
			where A, B are ions and x, y are coefficients...<br>
			<br>
			1. The relationship of the changes in amount (of which [[Mole (unit)\|mole]] is a unit), represented as N<sub>(∆)</sub>, between the species is given by [[Stoichiometry]] as follows:<br>
			<math>-\frac{N_{ AxBy(\Delta)}}{1} = \frac{N_{A(\Delta)}}{x} = \frac{N_{B(\Delta)}}{y}\,</math><br>
			which, when rearranged for ∆A and ∆B yields:<br>
			<math>N_{A(\Delta)} = -xN_{AxBy(\Delta)}\,</math><br>
			<math>N_{B(\Delta)} = -yN_{AxBy(\Delta)}\,</math><br>
			<br>

			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.<br>
			<math>N_{(i)} + N_{(\Delta)} = N_{(f)}\,</math> [[Difference operator\|The Difference law]]

			<math>N_{A(i)} = 0\,</math><br>
			<math>N_{B(i)} = 0\,</math>

			Which condense to the identities<br>
			<math>N_{A(f)} = N_{A(\Delta)}\,</math><br>
			<math>N_{B(f)} = N_{B(\Delta)}\,</math><br>
			<br>

			3. In these variables (with V for volume), the Molar solubility would be written as:<br>
			<math>S_0 = -\frac{N_{ AxBy(\Delta)}}{V}\,</math><br>
			<br>
			4. The [[Solubility Product]] expression is defined as:<br>
			<math>K_{sp} = [A]^x[B]^y\,</math><br>
			<br>


			These four sets of equations are enough to solve for S<sub>0</sub> [[algebra]]ically:<br>
			<br>
			<math>K_{sp} = {\left(\frac{N_{A(f)}}{V}\right)}^x {\left(\frac{N_{B(f)}}{V}\right)}^y\,</math>

			<math>K_{sp} = {\left(\frac{N_{A(\Delta)}}{V}\right)}^x {\left(\frac{N_{B(\Delta)}}{V}\right)}^y\,</math>

			<math>K_{sp} = \frac{{(N_{A(\Delta)})}^x {(N_{B(\Delta)})}^y}{V^{(x+y)}}\,</math>

			<math>K_{sp} = \frac{{(-xN_{AxBy(\Delta)})}^x {(-yN_{AxBy(\Delta)})}^y}{V^{(x+y)}}\,</math>

			<math>K_{sp} = \frac{{(-1)}^x {(x)}^x {(N_{AxBy(\Delta)})}^x {(-1)}^y {(y)}^y {(N_{AxBy(\Delta)})}^y}{V^{(x+y)}}\,</math>

			<math>K_{sp} = x^x y^y \frac{{(-1)}^x {(-1)}^y {(N_{AxBy(\Delta)})}^x {(N_{AxBy(\Delta)})}^y}{V^{(x+y)}}\,</math>

			<math>K_{sp} = x^x y^y \frac{{(-1)}^{(x+y)}{(N_{AxBy(\Delta)})}^{(x+y)}}{V^{(x+y)}}\,</math>

			<math>\frac{K_{sp}}{x^x y^y} = {\left(-\frac{N_{AxBy(\Delta)}}{V}\right)}^{(x+y)}\,</math>

			<math>\frac{K_{sp}}{x^x y^y} = {\left(S_0\right)}^{(x+y)}\,</math>

			Hence;

			<math>S_0 = \sqrt[(x+y)]{\frac{K_{sp}}{x^x y^y}}\,</math>

			==Simple calculation==
			If the [[solubility product]] constant (K<sub>sp</sub>) and dissociation product ions are known, the molar solubility can be computed without the aforementioned equation.

			===Example===
			Ionic substance AB<sub>2</sub> dissociates into A and 2B, or one mole of ion A and two moles of ion B. The soluble ion dissociation equation can thus be written as:

			AB<sub>2</sub> --> A + 2B

			where the corresponding solubility product equation is:

			K<sub>sp</sub> = [A][B]^2

			If the initial concentration of A is x, then that of B must be 2x. Inserting these initial concentrations into the solubility product equation gives

			K<sub>sp</sub> = (x)(2x)^2

			If Ksp is known, x can be computed, which is the molar solubility.

			{{DEFAULTSORT:Molar Solubility}}
			[[Category:Solutions]]

en>Andreas Sihono at 08:12, 30 August 2012

2012-08-30T08:12:45Z

New page

Nice to satisfy you, I am Marvella Shryock. Supervising is my occupation. One of the issues she enjoys most is to study comics and she'll be starting something else alongside with it. Puerto Rico is exactly where he's always been residing but she requirements to transfer because of her family.<br><br>Here is my web-site at home std testing ([http://Www.Revleft.com/vb/member.php?u=160656 More])