A 40% (w/v) aqueous stock solution of acetic acid is available. How would you prepare 250 ml of a 2.25% sol.?
143234
I think you at least need to know the density of pure acetic acid to solve your problem. We can assume density of water at 1.0 g/mL.\nIf we assume that the density of your 40% w/v solution is 1.0 g/mL then:\n\n    V = volume of 40% solution required would be: \n\n        V x .40  = 0.0225 x 250 \n\n  V = 14.1 mL \n\nYou place the 14.1 mL in a flask and add water to complete 250 mL and voila... You are done. \n\nLet me know if this is wrong.
365444
0.500L x 0.250moles/L x 58.44g/mol = 7.31g\n(a)
777607
Since they are both strong acids and bases you can assume: \n\nMaVa = MbVb\n\n(25/1000) * (1.2) = Mols NaOH\n\nconvert Mols to grams and that's your answer
169137
We have 48 gallons of a solution 64% acid\nSo the acid quantity = 48*0.64\n\nWe will add X gallons of a solution 16% acid\nSo the acid quantity = X*0.16\n\nThe Total acid quantity= 48*0.64 + X*.016\nThe Total Volume = 48 gallons + X gallons\n \nThe Average Percantage of acid= Total acid quantity/total volume\n=[48*0.64 + X*0.16]/[48+X] ===== 0.52 as given\n\n48*0.64 + X*0.16 = 0.52*(48+X)\n48*(0.64 - 0.52) = (0.52 - 0.16)*X\n48*0.12 = 0.36*X\nX=48 * 0.12/0.36= 48/3 = 16 gallons\n\nWe must add 16 gallons of solution 16% acid
780066
V = nRT/P\nV = (25g/44g/mol)(0.08206Latm/molK)(298K)/1atm\nV = 138.9 L
1145574
0.150 M NaOH\n\n0.150mol / L  X  40 g / mol  X  1 L / 1000 mL  X  200 mL  =  1.2 grams of solute (NaOH)
952447
I think you probably mean that the half liter is mixed with 250 ml. Otherwise the volume afterwards couldn't be 750 ml. Okay, the amount of HCl in the half liter is\nn=V*c=0.5l*2.5 mole/l=1.25 mole (remember that M simply means mole per liter).\nThe amount of HCl in the 250 ml is\nn=V*c=0.25l*3.75 mole/l=0.9375 mole\nThe total amount of HCl is then 1.25 mole+0.9375 mole=2.1875 mole meaning that the concentration in the final solution is\nc=n/V=2.1875 mole/0.750 l=2.92 mole/L=2.92 M.\n\nSince HCl is a strong acid it is completely separated in H+ and Cl- in the solution meaning that\npH=-log(concentration of H+)=-log(concentration of HCl)=-log(2.92)=-0.5.\n(Strictly speaking this formula for pH is only valid in dillute solutions like 0.1 M and below, but it is the best estimate we can give without making this problem VERY compliated).
271476
pH = 4.75\n\ntechnically, the pH can be anything.  But, for the most effective acetic acid buffer, the pH would be 4.75 because the pKa of acetic acid is 4.75.  When the pH = pKa, the amount of acid and base are equal, thus its resistant to pH changes is greatest.
607804
The easiest way to prepare an insoluble salt is to produce it from two aqueous solutions each containing an ion present in the salt!\n\nSo, you'll just mix two solutions, and the insoluble salt appears! Of course, if you intend to prepare a certain amount/ mass of the salt, you may need to carry out some calculations to determine the necessary concentrations/ volumes of the 2 initial solutions. \n\nYou'll also get another product in the reaction, but it should be soluble and will not contaminate the desired product (the insoluble salt).  \n\nYou may refer to this website to gain a better understanding...\nhttp://www.wpbschoolhouse.btinternet.co.uk/page01/AqueousChem/AqueousChem.htm#salts [Refer to point (4)]
754271
2H2 + O2 --> 2H2O\nPV = nRT\nP = 2 + 3 = 5\nT = 25 + 273 = 298\n5.00 atm x 55.0 L = n x .08206 x 298\nn = 11.25 mol H2 and O2 mixture\n11.25 mol x (2 mol H2O / 3 mol) = 7.50 mol H2O\n\nP x 55.0 = 7.50 x .08206 x 298\nP = 3.33 atm
955333
an easy example you can remember is H2O(water).\nif u want a more difficult one, u can use CH3COOH(acetic acid)!\n\nALL DE BEST!
745889
We can solve this using a system of equations.\n\nLet x be the volume of the 25% solution. Let y be the volume of the 60% solution.\n\nThe sums of the volumes must total 140 ml:\n\nx + y = 140.\n\nThe mass of CuS04 in the desired solution must equal the amounts added through the use of both solutions.\n\n30 percent of 140ml gives 42 ml of "pure" copper sulfate solution.\n\nThe 25% solution would contribute .25x of "pure" copper sulfate solution. The 60% solution would contribute .60y of "pure" copper sulfate solution, so:\n\n0.25x + 0.60y = 42.\n\nHere is the system:\n\nx + y = 140\n0.25x + 0.60y = 42\n\nThere are many different ways to solve the system. I'll use substitution here:\n\nx + y = 140, so y = 140 - x\n\n0.25x + 0.6(140 - x) = 42\n0.25x + 84 - 0.6x = 42\n-0.35x = -42\nx = 120\ny = 20\n\nSince x is the amount of 25% solution and y is the amount of 60% solution, we need 120 ml of 25% solution and 20 ml of 60% solution.\n\nThis appears to match your answer!
1072536
The liquid you want is a solution of boracic (or boric) acid.
985910
You have 2 38.0 mL volumes, so the total volume is 76.0 mL.\nThe reaction is H2SO4 + 2 KOH --> 2 H2O + K2SO4\nThe rise in temperature (delta T) is 30.17°C - 23.50°C = 6.67°C\n\nThe heat capacity of water is 1.0 cal/(g*°C)\nThe density of water is 1.0 g/mL\n\nThe H of the reaction is 1.0 cal/(g*°C) * 1.0 g/mL * 76.0 mL * 6.67°C = 506.9 cal or (506.9 cal * 4.186 J/cal * 0.001 kJ/J) = 2.122 kJ\n\nThere were 0.82 mol/L of KOH, and 38.0 mL, so there were .82*.038 mol = 0.03116 mol of KOH\n\nDeltaH per mole of KOH is 2.122 kJ / 0.03116 mol = 68.1 kJ/mol
792017
H2C3H2O4 + 2NaOH --> Na2C3H2O4 + 2H2O\n\nmole of NaOH needed = 0.1 * 20.76 = 2.76 mmole\nmole of malonic acid = 1.38 mmole\nmolarity of malonic acid = 1.38 / 13.15 = 0.104 M\n\npeace\nvixklen
760563
N=aM where a is the valency (depending on the type of chemical and reaction, e.g. number of H+/OH- for acid/bases per molecule, number of electrons gained/lost in redox reactions).\n\nFor AgNO3 a=1 (it is Ag+ so for redox or precipitation reactions it gives one equivalent group per molecule)\n\nSo M= N/a = N/1 =N\n\n1 M is 169.87 g/L =16.987g/100mL =16.987%\n\nUse M1*V1=M2*V2\n16.987*V1=1*V2\nV1 is the volume in mL of the stock solution that you will use to prepare V2 mL of 1%\nYou didn't specify how much you want to prepare...\n\nAlternatively you could have converted 1% to mole/L and do the math with molarities.
1140664
You're missing a unit. Do you mean 2.5 kg solution?