what is the normality of a solution that has a concentration of 106.41g/dL of H2SO4?
104563
My answer is:\n\n.217N
479787
In 100g of solution that is 5% by mass NaCl, there are 95 g H2O and 5g Nacl.\nYou want 250g of solution...set up a simple ratio:\n95g H2O/100g Soln = ?g H2O/250g Soln\n? = 237.5g H2O
867590
Well, for a start your equation shows copper(II) sulphide, not sulphate, and your equation is wrong even then.  That makes a bit of a difference!\n\nLet us imagine that you've got the formula right, but the name wrong.\n\nThe equation is simply Cu + S ---> CuS\n\nThen 64g will give (64+32)g.\n\nThus 64g will give 96g.\n\nSo 64 x (200/96)g of Cu will give 200g of product.
529624
its dilute h2so4 (38%) sulphuric acid.
324250
concentrated H2So4 is pretty bad
820326
If this is in Normality;\n\n0.1 Normal solution of  NaOH\n2 grams of NaOH  in a liter = 0.05 N\n\nN = weight in grams divided by MW (molecular weight ) = 40 / valence =1 / 1 liter. = 0.05\n\nMW = (Sodium) 23 + 16 (Oxygen) + 1 (hydrogen) = 40\n\nTherefore: Using the formula N1V1 = N2V2\n(0.1)( X) = (0.05) ( 1 liter)\n\nBy cross multiplication:\n\nX = (.0.05) ( 1) divided by ( 0.1) \nX =  0.5 liters or 500 milliliters\n\nbecause  1,000 ml = 1 liter\n\nYou will get the same answer for Molarity because the valence is one.\n\nso the answer is 500 milliliters.
109301
Normality generally refers to molecular weight of asa substance in grams dissolved in 1 litre of solvent.\nIn the present case 3.65 grams of HCl dissolved in 1 litre of water.\n\nMolarity refers to molecular weight of a substance dissolved in gram molecular weight of solvent.\nIn the present case 3.65 gram of HCl dissolved in 18 gram of water.
1171925
PH of the aqueous solution of NaCl is 6 - 7 @ 1% by weight
842422
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
738625
CuSO4+5H2O = 249.5 (calculate the RMM)\n\nSO4= 96\n\n96/249.5 x 100% = 38.48% of SO4
982757
I'm assuming that you're after a 10% solution of CuSO4 by weight.  \n\n1)  First find the mass of the H2O2 solution.\nYou can find the mass of the H2O2 solution because you know that the density of water is 1.0 g/mL and the solution is mostly water.  It doesn't matter that the solution is 3% H2O2 - they are only trying to throw you off.  The point here, is that the solution is mostly water and will have about the same mass as water.\n\nmass of H2O2 solution = \n10 mL H2O2 solution * 1 g H2O2 solution/1 mL of solution\n\nAnswer:  You have 10 g of H2O2 solution.\n\n2)  Now you can find the mass of CuSO4 that you need in order to get a 10% solution of CuSO4.  Remember that 10% literally means "10 grams CuSO4 per 100 grams Total".\n\nWhere:\nTotal = mass of CuSO4 + mass of H2O2 solution.\n\nObviously, you aren't going to make 100 grams, but the ratio still holds.  You can use the "10 g CuSO4 per 100 g Total" as a ratio with which to solve the equation.\n\nLets assign X = mass of CuSO4 you need.\n\nnow you know that X = 10% of Total \n\nThis is the same as saying 0.1 * Total\n\nand remember that:\nTotal = mass of CuSO4 + mass of H2O2 solution\n\nSo Total = X + mass of H2O2 solution\n\nYou also know that mass of H2O2 solution = 10 grams.\n\nSo here is the equation you must solve:\n\nX = 0.1 * (X + 10 grams)\n\nThe equation above literally reads, "The mass of CuSO4 equals 10 percent of the Total Mass.\n\nNow let's solve for X by distributing the right side using the informaton you already know:\n\nX = 0.1 * (X + 10 grams)\ndistribute the right side of the equation so:\n\nX = 0.1X + 1 gram\n\nSubtract 0.1X from both sides to isolate X:\n\nX - 0.1X = 0.1X - 0.1X + 1 gram\n\nSimplify\n\n0.9X = 1 gram\n\nDivide both sides by 0.9 to solve for X:\n\n0.9X/0.9 = 1 gram/0.9\n\nX = about 1.11 grams of CuSO4 needed.
35280
The Ksp of the reaction\n\nHgS <--> Hg + S\n\nis 2*10^-53. Extremely insoluble.
327703
The amount is diluted by a factor of 25.  So the resulting solution would have a concentration of 11.6M divided by 25 or \nor 0.464M.
940690
that would depend on the molarity of the substance. there could be any number of moles in that much solution.
424592
H2SO4 + 2NaOH ----> 2H2O + Na2SO4\n\nYOU MUST SEPARATE AND CANCEL THE COMMON IONS (AQUEOUS ONES)
795448
normality = molarity x ionizable H\n\n18 M x 2 H = 36 N
168718
The eq constant is 5.286E-16, so this means that 5.287E-16 = [H][OH]. (The [H2O] doesn't matter as it is a pure liquid.) The H and OH concentrations are the same (because of mole ratios), so solve this equation:\n\nx^2 = 5.287E-16\n\nx is the eq. concentration of H, and also the eq. concentration of OH.