If \(Q > K\), then the reactions shifts to the left to reach equilibrium, If \(Q < K\), then the reactions shifts to the right to reach equilibrium, If \(Q = K\) then the reaction is at equilibrium. The new expression would be written as: \[K'= \dfrac{1}{\dfrac{[G]^g[H]^h}{[A]^a[B]^b}} = \dfrac{[A]^a[B]^b}{[G]^g[H]^h}\]. Because the initial concentration of ethylene (0.155 M) is less than the concentration of hydrogen (0.200 M), ethylene is the limiting reactant; that is, no more than 0.155 M ethane can be formed from 0.155 M ethylene. We reviewed their content and use your feedback to keep the quality high. At equilibrium, a mixture of n-butane and isobutane at room temperature was found to contain 0.041 M isobutane and 0.016 M n-butane. The equilibrium mixture contained. Posted 7 years ago. \(P_{NO}=2x \; atm=1.8 \times 10^{16} \;atm \). Given: balanced chemical equation, \(K\), and initial concentrations of reactants. or neither? { "15.01:_The_Concept_of_Dynamic_Equilibrium" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "15.02:_The_Equilibrium_Constant_(K)" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "15.03:_Expressing_the_Equilibrium_Constant_in_Terms_of_Pressure" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "15.04:_Calculating_the_Equilibrium_Constant_From_Measured_Equilibrium_Concentrations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "15.05:_Heterogenous_Equilibria_-_Reactions_Involving_Solids_and_Liquids" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "15.06:_The_Reaction_Quotient-_Predicting_the_Direction_of_Change" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "15.07:_Finding_Equilibrium_Concentrations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "15.08:_Le_Chateliers_Principle-_How_a_System_at_Equilibrium_Responds_to_Disturbances" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "01:_Matter_Measurement_and_Problem_Solving" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "02:_Atoms_and_Elements" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "03:_Molecules_Compounds_and_Chemical_Equations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "04:_Chemical_Reactions_and_Aqueous_Reactions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "05:_Gases" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "06:_Thermochemistry" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "07:_The_Quantum-Mechanical_Model_of_the_Atom" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "08:_Periodic_Properties_of_the_Elements" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "09:_Chemical_Bonding_I-_Lewis_Structures_and_Determining_Molecular_Shapes" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "10:_Chemical_Bonding_II-_Valance_Bond_Theory_and_Molecular_Orbital_Theory" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "11:_Liquids_Solids_and_Intermolecular_Forces" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "12:_Solids_and_Modern_Materials" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "13:_Solutions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "14:_Chemical_Kinetics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "15:_Chemical_Equilibrium" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16:_Acids_and_Bases" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "17:_Aqueous_Ionic_Equilibrium" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "18:_Gibbs_Energy_and_Thermodynamics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "19:_Electrochemistry" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "20:_Radioactivity_and_Nuclear_Chemistry" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "21:_Organic_Chemistry" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "22:_Biochemistry" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "23:_Chemistry_of_the_Nonmetals" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "24:_Metals_and_Metallurgy" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "25:_Transition_Metals_and_Coordination_Compounds" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, [ "article:topic", "showtoc:no", "license:ccbyncsa", "transcluded:yes", "licenseversion:40" ], https://chem.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FBookshelves%2FGeneral_Chemistry%2FMap%253A_A_Molecular_Approach_(Tro)%2F15%253A_Chemical_Equilibrium%2F15.07%253A_Finding_Equilibrium_Concentrations, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), Example \(\PageIndex{3}\): The watergas shift reaction, 15.6: The Reaction Quotient- Predicting the Direction of Change, 15.8: Le Chteliers Principle- How a System at Equilibrium Responds to Disturbances, Calculating an Equilibrium Constant from Equilibrium Concentrations, Calculating Equilibrium Concentrations from the Equilibrium Constant, Using ICE Tables to find Kc(opens in new window), Using ICE Tables to find Eq. \(K = 0.106\) at 700 K. If a mixture of gases that initially contains 0.0150 M \(H_2\) and 0.0150 M \(CO_2\) is allowed to equilibrate at 700 K, what are the final concentrations of all substances present? The Equilibrium Constant is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts. In the section "Visualizing Q," the initial values of Q depend on whether initially the reaction is all products, or all reactants. We could solve this equation with the quadratic formula, but it is far easier to solve for \(x\) by recognizing that the left side of the equation is a perfect square; that is, \[\dfrac{x^2}{(0.0150x)^2}=\left(\dfrac{x}{0.0150x}\right)^2=0.106\nonumber \]. From a mathematical perspective, with the activities of solids and liquids and solvents equal one, these substances do not affect the overall K or Q value. By calculating Q (products/reactants), you can compare it to the K value (products/reactants AT EQUILIBRIUM) to see if the reaction is at equilibrium or not. Use the coefficients in the balanced chemical equation to obtain the changes in concentration of all other substances in the reaction. Write the equilibrium constant expression for the reaction. To solve quantitative problems involving chemical equilibriums. Direct link to Vedant Walia's post why shouldn't K or Q cont, Posted 7 years ago. When can we make such an assumption? Check your answers by substituting these values into the equilibrium constant expression to obtain \(K\). Equilibrium constant are actually defined using activities, not concentrations. How can we identify products and reactants? We can now calculate the equilibrium constant for the reaction: \[K=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}=\dfrac{(0.056)^2(0.028)}{(0.444)^2}=4.5 \times 10^{4}\nonumber \], The German chemist Fritz Haber (18681934; Nobel Prize in Chemistry 1918) was able to synthesize ammonia (\(NH_3\)) by reacting \(0.1248\; M \;H_2\) and \(0.0416\; M \;N_2\) at about 500C. http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/ICEchart.htm. (Remember that equilibrium constants are unitless.). Solution Direct link to Everett Ziegenfuss's post Would adding excess react, Posted 7 years ago. Because \(K\) is essentially the same as the value given in the problem, our calculations are confirmed. We obtain the final concentrations by substituting this \(x\) value into the expressions for the final concentrations of n-butane and isobutane listed in the table: \[[\text{n-butane}]_f = (1.00 x) M = (1.00 0.72) M = 0.28\; M \nonumber \], \[[\text{isobutane}]_f = (0.00 + x) M = (0.00 + 0.72) M = 0.72\; M \nonumber \]. Direct link to rbrtweigel's post K is the equilibrium cons, Posted 8 years ago. In many situations it is not necessary to solve a quadratic (or higher-order) equation. Direct link to Ibeh JohnMark Somtochukwu's post the reaction quotient is , Posted 7 years ago. N 2 O 4 ( g) 2 NO 2 ( g) Solve for the equilibrium concentrations for each experiment (given in columns 4 and 5). We insert these values into the following table: C We use the stoichiometric relationships given in the balanced chemical equation to find the change in the concentration of \(Cl_2\), the substance for which initial and final concentrations are known: \[[Cl_2] = 0.028 \;M_{(final)} 0.00\; M_{(initial)}] = +0.028\; M\nonumber \]. At any given point, the reaction may or may not be at equilibrium. Calculate the final concentrations of all species present. Direct link to Rajnikant Roy's post How is the Reaction Const, Posted 3 years ago. Substituting these concentrations into the equilibrium constant expression, \[K=\dfrac{[\textit{isobutane}]}{[\textit{n-butane}]}=0.041\; M = 2.6 \label{Eq2} \]. At equilibrium, concentrations of all substances are constant. Various methods can be used to solve the two fundamental types of equilibrium problems: (1) those in which we calculate the concentrations of reactants and products at equilibrium and (2) those in which we use the equilibrium constant and the initial concentrations of reactants to determine the composition of the equilibrium mixture. Direct link to Srk's post If Q is not equal to Kc, , Posted 5 years ago. What is the composition of the reaction mixture at equilibrium? I get that the equilibrium constant changes with temperature. \([H_2]_f=[H_2]_i+[H_2]=(0.01500.00369) \;M=0.0113\; M\), \([CO_2]_f =[CO_2]_i+[CO_2]=(0.01500.00369)\; M=0.0113\; M\), \([H_2O]_f=[H_2O]_i+[H_2O]=(0+0.00369) \;M=0.00369\; M\), \([CO]_f=[CO]_i+[CO]=(0+0.00369)\; M=0.00369 \;M\). Hydrogen gas reacts with iodine vapor to give hydrogen iodide according to the following chemical equation: \(K = 54\) at 425C. The yellowish sand is covered with people on beach towels, and there are also some swimmers in the blue-green ocean. The problem then is identical to that in Example \(\PageIndex{5}\). For example, 1 mol of \(CO\) is produced for every 1 mol of \(H_2O\), so the change in the \(CO\) concentration can be expressed as \([CO] = +x\). What is \(K\) for the reaction, \[N_2+3H_2 \rightleftharpoons 2NH_3\nonumber \], \(K = 0.105\) and \(K_p = 2.61 \times 10^{-5}\), A Video Disucssing Using ICE Tables to find Kc: Using ICE Tables to find Kc(opens in new window) [youtu.be]. when setting up an ICE chart where and how do you decide which will be -x and which will be x? Initial reactant and product concentrations and equilibrium concentrations (in M) are given as well as the equilibrium constants (at 25 C). Say if I had H2O (g) as either the product or reactant. If a mixture of 0.257 M \(H_2\) and 0.392 M \(Cl_2\) is allowed to equilibrate at 47C, what is the equilibrium composition of the mixture? Construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations (as initial concentrations plus changes in concentrations). What is the partial pressure of NO in equilibrium with \(N_2\) and \(O_2\) in the atmosphere (at 1 atm, \(P_{N_2} = 0.78\; atm\) and \(P_{O_2} = 0.21\; atm\)? We can then express the final concentrations in terms of the initial concentrations and the changes they have undergone. Then substitute values from the table into the expression to solve for \(x\) (the change in concentration). B. If the product of the reaction is a solvent, the numerator equals one, which is illustrated in the following reaction: \[ H^+_{(aq)} + OH^_{(aq)} \rightarrow H_2O_{ (l)}\]. H. When we aren't sure if our reaction is at equilibrium, we can calculate the reaction quotient, At this point, you might be wondering why this equation looks so familiar and how. Legal. Activity is expressed by the dimensionless ratio \(\frac{[X]}{c^{\circ}}\) where \([X]\) signifies the molarity of the molecule and c is the chosen reference state: For gases that do not follow the ideal gas laws, using activities will accurately determine the equilibrium constant that changes when concentration or pressure varies. \([C_2H_6]_f = (0.155 x)\; M = 0.155 \; M\), \([C_2H_4]_f = x\; M = 3.6 \times 10^{19} M \), \([H_2]_f = (0.045 + x) \;M = 0.045 \;M\). Construct a table showing what is known and what needs to be calculated. with \(K_p = 4.0 \times 10^{31}\) at 47C. In this case, the equation is already balanced, and the equilibrium constant expression is as follows: \[K=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\nonumber \]. The reaction is already at equilibrium! To simplify things a bit, the line can be roughly divided into three regions. If this assumption is correct, then to two significant figures, \((0.78 x) = 0.78\) and \((0.21 x) = 0.21\). Concentrations & Kc: Using ICE Tables to find Eq. The colors vary, with the leftmost vial frosted over and colorless and the second vial to the left containing a dark yellow liquid and gas. \(2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} \), \(N_2O_{ (g)} + \dfrac{1}{2} O_{2(g)} \rightleftharpoons 2NO_{(g)} \), \(Cu_{(s)} + 2Ag^+_{(aq)} \rightleftharpoons Cu^{+2}_{(aq)} + 2Ag_{(s)} \), \(CaCO_{3 (g)} \rightleftharpoons CaCO_{(s)} + CO_{2 (g)} \), \(2NaHCO_{3 (s)} \rightleftharpoons Na_2CO_{3 (s)} + CO_{2 (g)} + H_2O_{ (g) }\). The final equilibrium concentrations are the sums of the concentrations for the forward and reverse reactions. Concentrations & Kc(opens in new window) [youtu.be]. Consider the following reaction: H 2O + CO H 2 + CO 2 Suppose you were to start the reaction with some amount of each reactant (and no H 2 or CO 2). The equilibrium constant is a ratio of the concentration of the products to the concentration of the reactants. Let's take a look at the equilibrium reaction that takes place between sulfur dioxide and oxygen to produce sulfur trioxide: The reaction is at equilibrium at some temperature. in the example shown, I'm a little confused as to how the 15M from the products was calculated. Example \(\PageIndex{3}\) illustrates a common type of equilibrium problem that you are likely to encounter. Direct link to Priyanka Shingrani's post in the above example how , Posted 7 years ago. The final \(K_p\) agrees with the value given at the beginning of this example. C The final concentrations of all species in the reaction mixture are as follows: We can check our work by inserting the calculated values back into the equilibrium constant expression: \[K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(0.00369)^2}{(0.0113)^2}=0.107\nonumber \]. C) The rate of the reaction in the forward direction is equal to the rate of the reaction in the reverse direction. Check out 'Buffers, Titrations, and Solubility Equilibria'. The activity of pure liquids and solids is 1 and the activity of a solution can be estimated using its concentration. It's important to emphasize that chemical equilibria are dynamic; a reaction at . This equation can be solved using the quadratic formula: \[ x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{0.127 \pm \sqrt{(0.127)^24(0.894)(0.0382)}}{2(0.894)}\nonumber \], \[x =0.148 \text{ and } 0.290\nonumber \]. Direct link to yuki's post We didn't calculate that,, Posted 7 years ago. The concentration of nitrogen dioxide starts at zero and increases until it stays constant at the equilibrium concentration. Any videos or areas using this information with the ICE theory? The equilibrium constant expression is written as follows: \[K_c = \dfrac{[G]^g[H]^h}{1 \times 1} = [G]^g[H]^h\]. Explanation: Advertisement 2.59 x 10^24 atoms of Ga = ___mol Ga the reaction quotient is affected by factors just the same way it affects the rate of reaction. If Q is not equal to Kc, then the reaction is not occurring at the Standard Conditions of the reaction. The initial partial pressure of \(O_2\) is 0.21 atm and that of \(N_2\) is 0.78 atm. of the reactants. The chemical equation for the reaction of hydrogen with ethylene (\(C_2H_4\)) to give ethane (\(C_2H_6\)) is as follows: \[H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)}\nonumber \]. \([H_2]_f = 4.8 \times 10^{32}\; M\) \([Cl_2]_f = 0.135\; M\) \([HCl]_f = 0.514\; M\), A Video Discussing Using ICE Tables to find Eq. If Kc is larger than 1 it would mean that the equilibrium is starting to favour the products however it doesnt necessarily mean that that the molar concentration of reactants is negligible. Under these conditions, there is usually no way to simplify the problem, and we must determine the equilibrium concentrations with other means. I don't get how it changes with temperature. at equilibrium. The reaction quotient is calculated the same way as is \(K\), but is not necessarily equal to \(K\). Conversely, removal of some of the reactants or products will result in the reaction moving in the direction that forms more of what was removed. Taking the square root of the middle and right terms, \[\dfrac{x}{(0.0150x)} =(0.106)^{1/2}=0.326\nonumber \], \[x =0.00369=3.69 \times 10^{3}\nonumber \]. For more information on equilibrium constant expressions please visit the Wikipedia site: The image below can be found here: image.tutorvista.com/content/chemical-equilibrium/reaction-rate-time-graph.gif, \(K_c = \dfrac{ [SO_3]^2}{[O_2][SO_2]^2}\), \(Kc = \dfrac{ [NO]^2}{[O_2]^{0.5}[N_2O]}\), \( H_2 (g)+ I_2 (g) \rightarrow 2HI(g) \), Modified by Tom Neils (Grand Rapids Community College). If Q=K, the reaction is at equilibrium. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Direct link to Lily Martin's post why aren't pure liquids a, Posted 6 years ago. A reversible reaction can proceed in both the forward and backward directions. Example 10.3.4 Determine the value of K for the reaction SO 2(g) + NO 2(g) SO 3(g) + NO(g) when the equilibrium concentrations are: [SO 2] = 1.20M, [NO 2] = 0.60M, [NO] = 1.6M, and [SO 3] = 2.2M. This is because the balanced chemical equation for the reaction tells us that 1 mol of n-butane is consumed for every 1 mol of isobutane produced. B We can now use the equilibrium equation and the known \(K\) value to solve for \(x\): \[K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{x^2}{(0.570x)(0.632x)}=0.106\nonumber \]. When an equilibrium constant is calculated from equilibrium concentrations, molar concentrations or partial pressures are substituted into the equilibrium constant expression for the reaction. C The small \(x\) value indicates that our assumption concerning the reverse reaction is correct, and we can therefore calculate the final concentrations by evaluating the expressions from the last line of the table: We can verify our calculations by substituting the final concentrations into the equilibrium constant expression: \[K=\dfrac{[C_2H_6]}{[H_2][C_2H_4]}=\dfrac{0.155}{(0.045)(3.6 \times 10^{19})}=9.6 \times 10^{18}\nonumber \]. YES! In this section, we describe methods for solving both kinds of problems. It is used to determine which way the reaction will proceed at any given point in time. A Because we are given Kp and partial pressures are reported in atmospheres, we will use partial pressures. open bracket, start text, N, O, end text, close bracket, squared, equals, K, open bracket, start text, N, end text, start subscript, 2, end subscript, close bracket, open bracket, start text, O, end text, start subscript, 2, end subscript, close bracket, space, space, space, space, space, space, space, start text, T, a, k, e, space, t, h, e, space, s, q, u, a, r, e, space, r, o, o, t, space, o, f, space, b, o, t, h, space, s, i, d, e, s, space, t, o, space, s, o, l, v, e, space, f, o, r, space, open bracket, N, O, close bracket, point, end text, open bracket, start text, N, O, end text, close bracket, equals, square root of, K, open bracket, start text, N, end text, start subscript, 2, end subscript, close bracket, open bracket, start text, O, end text, start subscript, 2, end subscript, close bracket, end square root, 5, point, 8, times, 10, start superscript, minus, 12, end superscript, start text, M, end text, open bracket, start text, N, end text, start subscript, 2, end subscript, close bracket, open bracket, start text, O, end text, start subscript, 2, end subscript, close bracket.
Edinburgh Raaf Base Medical Centre,
Alamat Ng Rosas Tauhan,
South Dakota Utv Width Restrictions,
Articles A
