how much concentrated 18m sulfuric acid is needed to prepare 250ml of a 6.0 m solution?
493729
almost right ...\n\nc1* v1 = c2 * v2\n18M (V1) = 6M*0.25L\n\nBut that makes V1 = 0.083L
1012209
300mL\n    x.18\n_______\n\n   54.00 mL
16164
The instruction is not correct because the total volume yielded when you mix 100ml of concentrated solution and 900ml water will be somewhat less than 1000ml. The concentration will be slightly higher than 1.0M\n\nThe correct procedure is to take a 1l volumetric flask  and add the concentrated solution to about 800ml water, then fill up to the mark.
940690
that would depend on the molarity of the substance. there could be any number of moles in that much solution.
215604
C6H12O6(s) ---H2SO4--> 6C(s) + 6H2O(in H2SO4 solu.) + Heat
884562
Doesn't matter how much you started with.  The final amount remaining in the water at 20 will be 0.22g/100mL of solvent.\n\nSo your system will deposit 1.53g of SA at 20C.
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.
683322
Well, I suppose 1.25 g is 1250 mg, so seeing there's 1250 mg in each ml of solution, taking out 0.2 ml of the solution will net you 250 mg of the substance.
738625
CuSO4+5H2O = 249.5 (calculate the RMM)\n\nSO4= 96\n\n96/249.5 x 100% = 38.48% of SO4
906606
1) To prepare this solution you should do:\n\n100 g of solution needs 2.5 g of solute (to get 2.5% mass/mass)\n150 g of solution needs  x    g of solute (to get 2.5% mass/mass)\n\n2)  x = (150 g x 2.5 g / 100 g) = 3.75 grames of KCl\n\nThat's it!\n\nGood luck!
320061
Someone else has asked this same question before, and it really isn't that hard . . .\n\n60 ml of 0.250M HCl plus 60 ml of 0.125M NaOH will result in 120 ml of solution total.  All of the NaOH will have been neutralized by the HCl, leaving an excess of HCl, so the solution will be acidic -- the pH will be much less than 7.\n\n60 ml * (1.0 liter/1000 ml) * (0.250 mole/liter HCl) = .015 mole HCl\n\n60 ml * (1.0 liter/1000 ml) * (0.125 mole/liter NaOH) = .0075 moles NaOH\n\nThat means you will have 0.015 - 0.075 = 0.075 moles of HCl left over after the neutralization.\n\nI leave it for you to calculate the pH of a solution of 0.075 moles of HCl in 120 ml of solution.  Remember that pH = -log(base 10)[H+]
105753
First you must find the number of moles of 12.6 Liters of SO2 gas at STP.\n\nAt STP an ideal gas (we will assume SO2 is an ideal gas) has a volume of 22.4 Liters.\nSo how many moles would be in 12.6 Liters?\n\nAfter you obtain the number of moles of SO2, then figure out how many moles of Sulfur there are in 1 mole of SO2.\nFor every 1 molecule of SO2, there is 1 Sulfur atom, so for every 1 mole SO2, there is 1 mole of Sulfur.  Knowing this, you can calculate the number of moles of Sulfur needed.\n\nBy multiplying tbe number of moles of Sulfur needed by the molar mass of Sulfur (32.07 g/mol) you can find the mass of the Sulfur required.
877763
OK, I think I have this one pegged:\nThe original mixture is 10% acid and 90% water and the final mixture is 60% acid and 40% water after pouring some amount of the original mixture into 5 liters of pure 100% acid. Set up two equations:\nfor the acid: x (10) + 5 (100) = y (60) where x = original mixture volume and y = volume of resulting mixture.\nfor the water: x (90) + 0 (100) = y (40)\nthen these can be reduced to:\n10x + 500 = 60 y     and  90x = 40y\nsolve the simultaneous equations to find x and y\nThe volume of water in the final mix is equal to y times .40\nGood luck
810972
Cool stuff, I used to get confused by ugly chemistry calculations, too :D\n\nYou have 3.88 g acid in 250cm3, so in the 25 cm3 you titrated there are 0.388 g of the acid. You need to know how many moles that is, so that you can calculate molar mass.\n\nThe amount of NaOH used up is 46.5 cm3 × 0.095 mol/dm3, that is 0.0465 dm3 × 0.095 M, which equals 0.00442 mol (It's not exactly that much, but I'll stick with three significant figures).\n\nSince NaOH reacts with monoprotic acids in a 1:1 ratio, that means that there was exactly the same amount of acid in the solution.\n\nNow we know that 0.388 g of the acid is 0.00442 mol, so we can divide the mass by the number of moles to get molar mass. That is 0.388g/0.00442 mol = 87.8 g/mol.\n\nRelative molecular mass is simply molar mass in grams, but without the unit, so Mr of the acid is 87.8.
758773
6O2 + C6H12O6 ---> 6CO2 + 6H2O
211855
it's about 6M sulfuric acid. 18M is almost an oil and is used mostlty for transporting.
1186593
Mix with a 0% solution in the relation 33:2\n\nSo use 2 units of your solution (like 2 liters) and 33 units of the 0% solution (like 33 liters).\n\nThat's all.