ChemLab: Beer's law: How do I find the concentration of a complex w/ a given stock solution?
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When you do dilutions you need to use M1V1=M2V2 to figure out the concentration of each solution.  You know how much of the stock solution was put into each, and the volume of stock, and the total volume so use algebra and figure out the concentration of the diluted solution.\n\nYou haven't gotten to the phase where you use Beer's law.  that comes after you test your dilutions and test an unknown.
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let me see if i can at least help you with part of your problem, solubility in chemistry is simple, basically the equation will always be:\n\n[Solubility equalibrium constant] = concentration of product divided by concentration of reagents (concentration of each reagents multiplying by each other).\n\nthe easiest way to understand this is to simplify things for you so as you know, a chemical equation always go like this:\n\nreagents <==> products\n\nJust know this from the above, the dynamic here is that when you have an excess of either the reagents or the product, the balance of the equation tend to shift in the opposite direction (ie excess concentration of product will cause more reagents to form and vice versa), in a chemical equation such as the one from above, the reaction can actually go either way, products or reagents can form.  What really matters is the equalibrium constant that I talked about in the first equation. This constant represent the concentration of the products and the reagents when they are in equailibrium or when they are balance, meaning that the rate at which products or reagent form stays the same and there are no change in the concentration of either the products or the reagents so the whole thing is static at the point.   Think about this, more concentration of products wouldn't form unless you have extra reagents to spare that can shift the balance.  When something is saturated it means that it can longer dissolve in a solvent (let's use water to simplify the explanation)so it just accumulates at the bottom, water molecules can only separate so many ions from each other (the reagents) and preventing them from forming a solid product since water itself has limited molecules (if you mix enough table salt in a glass of water, it'll get to the point where it'll accumulate at the bottom and won't dissolve).  If remember correctly, in a chemical equation, when there's different amount of reagents needed to form the product, there's a slight modification to the equation I gave you, for example:\n\n2 CaCO3 + 4 ZnO2 <----> 3 ZnCO3 + 3 CaO2\n\nthen  [equilibrium constant] = [Cao2]^3 x [ZnCO3]^3 divided by [CaCO3]^2 x [ZnO2]^2\n\nnote that you should balance the chemical equations first, I am just giving an example.\n\nhere's a summary of how to compare numbers, if you plug given concentrations of reagents and product into the equation and the number you get is larger than the real equilibrium constant, then the concentration of products is higher than it should be and the equation will shift towards forming more reagents (in the case of the solid product, it'll dissolve) and if there's more reagents, more solid product is gonna form.  Hope that helps. Remember the Chemistry textbook is still your best reference.
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Yes, it is true that the density of a solution is different of solvant but tjis difference is small ; So , i propose to dissove 40g of amyl acetate in a liter of water if that acetate is a solid. If that acetate is aliquid the total volume will be 1.04 l and you must use 47 cc tp water
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For the first question , the key words are FINAL MOLAR CONCENTRATION.  This assumes that the reaction has gone to completion and kinetics are no longer involved.  Since the balanced reaction is A + 2B ---> 3C, you can determine that the limited reagent is "B" and that "C" is created at a 2:3 ratio.  At the end of the reaction there will be 0.180M in "C" since is was created in a 2:3 ratio starting with 0.120M "B".\n\nAs for the second question, graphs can be used to determine if the reaction kinetics are first order or second order.  If you make a graph plotting the natural log of the pressure of "A" versus time ( ln(A) vs. t ), and you get a straight line graph, then the reaction will be exhibiting first order kinetics.  If you make a graph plotting the inverse of the pressure of "A" versus time ( 1/(A) vs. t ) and you get a straight line graph, then the reaction will be exhibiting second order kinetics.
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NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
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Themes > Science > Chemistry > Miscellenous > Help file Index > Precipitation Equilibrium > Ksp and Ion Concentration \n\nThe solubility product constant can be used to compute the equilibrium concentration of ions in solution in the same way that you use other equilibrium constant expressions to compute equilibrium concentrations. \n\nUse the Initial, Change, Equilibrium method to compute the concentrations: write the equilibrium expression, the initial amounts of each thing in the expression, how much they change by, then the equilibrium amounts. From that, you can compute the concentrations of all species. \n\nExample: BaF2 has a Ksp of 1.8*10-7. What is the concentration of barium in a saturated solution of BaF2? \n\nSolution:. First, write the balanced equation and the solubility constant expression: \n\nBaF2(s) < = > Ba+2(aq) + 2F-(aq) \nKsp = [Ba+2][F-]2\nWe start with no barium or fluoride ions in solution. If we set the amount of barium ions formed to x, then the number of flouride ions formed is just 2x \n\n [Ba+2] (M) [F-] (M) \nInitial 0 0 \nChange +x +2x \nEquilibrium x 2x \n\n\nNow plug in the equilibrium values shown above and solve for x \nKsp = [Ba+2][F-]2 \nKsp = (x)(2x)2 \nx = 3.6*10-3\nSince we defined the concentration of Ba+2 to be x, we've solved the problem \n[Ba+2] = 3.6*10-3 \n\nInformation provided by: http://learn.chem.vt.edu\n \n**try to visit the site below..** You can calculate the Ksp based on the above example..I'm sure the answer is in your books and notes..have fun learning..
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1. Determine the molar mass of KHC8H4O4 = 204.1 g/mol\n\n2. Determine the number of moles: \n(0.700g)/(204.1g/mol) = 0.00343 mol\n\n3. Determine the molarity of the three trials. This will show you the concentration of NaOH.\na.  0.00343 mol/ 0.0384L = 0.089 M\nb.  0.00343 mol/ 0.0370L = 0.093 M\nc.  0.00343 mol/ 0.0361L = 0.095 M\n\n4. Get the average of the concentration of the three trials: \n( 0.089 M + 0.093 M + 0.095 M)/3 = 0.092 M\n\nHence, the actual concentration of your 0.1M NaOH is 0.092M.
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Initial concentration B: 8/20=0.4 M\nInitial concentration C: 12/20= 0.6 M\n\n.. .. .. .. .. .. 2A <=> B + C\nInitial .. .. .. .. .. .. .. 0.4 .. 0.6\nReact .. .. .. .. .. .. .. x .. ..x\nProduce ..  2x\nAt Equil. .. . 2x .. ..0.4-x.. 0.6-x\n\nK=[B][C]/[A]^2 =>\n0.035=(0.4-x)(0.6-x)/(2x)^2 =>\n0.035*4x^2= 0.24 -0.4x -0.6x + x^2 =>\n0.86x^2 -x+0.24 =0\nIt is a quadratic equation of the type ax^2+bx+c=0\nThere are two possible solutions:\n\nx1= (-b+Squareroot( b^2-4ac))/2a\nx2= (-b-Squareroot( b^2-4ac))/2a\n\nx1 =(-(-1)+ SQRT( (-1)^2 -4*0.86*0.24) ) / (2*0.86)= 0.824\nx2=0.339\n\nx1 is rejected since it is bigger than the concentration of B which is the limiting reagent (and also bigger of C) and we can't have negative values for the equilibrium concentration of B (and C)\nThus the only acceptable solution is x=x2= 0.339\n[A]=2*x= 2*0.339= 0.678 M
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If the mole of HCl is unkonwn, u ll need a pH meter o something else to determine the pH of the solution first.\nBy using the formula:\npH = -log(concentration of H+)\nthen u can get the concentration of H+\nHCl is strong acid, so the concentration of H+ equals to the concentrtion of HCl
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Some quick help now\n1. Brass is a solid solution; copper is the solvent and zinc is the solute it is also an alloy so it may get you 6 as well. \n2. A bottle or can of a soft drink unopened is LeChatelier's Principle. It is at stasis until it is opened, thus the volume is larger and the equilibrium is disturbed until it gets back to equilibrium by outgassing the CO2.\n3. Take salt and sugar and mix them together. Large state of entropy. Hard to unmix them. \n5. Volatile solid: mothballs - naptha. Viscuous liquid - container of car engine oil\n7. That is a mole of something. Water H2O - 2 Hydrogens + 1 oxygen  (2 X 1.00 + 1X 15.999 = 18 atomic mass units) = 18 amus expressed as grams = 18 grams has 1 mole 18 grams of water is 18 ccs. So get something that holds 18 ccs or 18 mls of water. Small bottle.
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Try it yourself, its not that bad. \nHere is how:\n1) Set up a stoichometry ratio.\n2) Using the mass that they give you, calculate the theoretical yield of each product.\n3) Whichever product has a smaller theoretical yield, analyze the reactants which caused it to be such.\n4) The one that 'limits' the amount of products is the limiting reagent.\n5) Using that value, calcuate the amount of the other reactant that is necessary. Subtract from the orginal value to find the excess reactant.
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*your equation is written wrong it should be:\n3Cu(NO3)2 + 2Na3(PO4) -> Cu3(PO4)2 + 6Na(NO3)\n\nyou have to find your limiter and your excess.\n 1.) change grams to moles using moler mass\n2.2 g Cu(NO3)2 times 1mol/188 g = .0117mol\n1.29 Na3(PO4) times 1mol/164 g = .00787 mol\n\n2.) divide the two by the # of moles in the balanced equation:\nCu(NO3)2: .0117 mol/3 = .00390 mol\n Na3(PO4): .00787 mol/2=.00394 mol\n\n3.) So now you know you don't really have a limiter, but we'll go with Cu(NO3)2 b/c there's a little less. \n\nthere are 3 moles of Cu(NO3)2 used for every 1 mol of Cu3(PO4) produced.\n\n.0117 mol Cu(NO3)2 times (1 mol Cu3(PO4) / 3 mol Cu(NO3)2) = .00390 mol Cu3(PO4) produced\n\nconvert that to grams\n\n.00390 mol Cu3(PO4) times 286 g/mol = 1.12 g \n\nthat's your actual\n\nso % yield = .89/1.12 x 100 = 79.5%\n\noh yeah you find moler mass by adding the molecular weights of the atoms in the compound (times the number of the atoms in that compound) for ex. Cu3(PO4) 63.6 x 3 + 31.0 + 16.0 x 4 = 286\n\nDon't forget significant figures!!\n\nhope that helped!
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The first step, find the moles of both.\nYou should get 25 mmol benzoic acid, and 15.6 mmol NaOH.\n\nThen you must find after the reactions, how many moles remain.  \nYou should get 9.4 mmol benzoic acid, 0 moles NaOH, and 15.4 mmol benzoate.\n\nYou have some benzoic acid and some benzoate ion, a conjugate acid/base pair.  This is a buffer solution, so use the Henderson-Hesselbach equation.  You should get pH = 4.4.
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HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)
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Mstock*Vstock=Mdil*Vdil =>\n\nVstock=(Mdil/Mstock)*Vdil = (0.04/Mstock)*500 mL\n\nSo when they give you the molarity of the stock solution you just plug it in the equation and get the volume (in mL) of stock that you need to use.
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Chemisty, A Formula For Answers
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I assume the reaction in question is:\n\nA(g) <-> 2B(g)\n\nThe equilibrium constant, Keq, for this reaction is given by:\n\nKeq = ([B]^2)/[A]\n\nwhere the quantities in square brackets mean "the concentration of" the relevant species at equilibrium (or, in the case of gases, one commonly uses equilibrium partial pressures instead of concentrations).\n\nPlugging in the values we are given for the equilibrium concentrations, we have that:\n\nKeq = (0.154^2)/0.381\n\nKeq = 0.0622