if the drinking water standard for the sulfate ion is 250 ppm, this is equal to?
954132
Yes\nFor solutions \nparts per hundred= %= g/100ml\nso ppm= 10^-4* %= 10^-4 g/100mL= 10^-3 g/L= mg/L
529624
its dilute h2so4 (38%) sulphuric acid.
738625
CuSO4+5H2O = 249.5 (calculate the RMM)\n\nSO4= 96\n\n96/249.5 x 100% = 38.48% of SO4
438725
Magnesium Sulfate\n\nMgSO4\n\nMg(OH)2 + H2SO4 -> MgSO4 + 2HOH\n\nHOH = H20\nJust in case...
595177
Actually, this is pretty simple.  You have a saturated  solution of lead sulfate (PbSO4) and strontium sulfate (SrSO4).  Probably the biggest hurdle here is to realize that the concentration of sulfate ion will be the same for both salts!  Therefore, the concentration of sulfate ion will be the result of the most soluble of the two salts:\n\nKsp for PbSO4 = [Pb+2]*[SO4-2] = 1.6x10^-8\nKsp for SrSO4 = [Sr+2]*[SO4-2] = 2.8x10^-7\n\nStrontium sulfate is the more soluble of the two salts, so it will control the concentration of sulfate:\n\nSrSO4 ===> Sr+2  + SO4-2, thus [Sr+2] = [SO4-2]\n\nKsp for SrSO4 = 2.8x10^-7 = [SO4-2]^2 , so \n\n[SO4-2] = sqrt(2.8x10^-7) = 5.29x10^-4 = [Sr+2]\n\nNow, to find out the concentration of Pb+2, we use the value we have just calculated for [SO4-2] and the Ksp:\n\nKsp for PbSO4 = [Pb+2] * [SO4-2] = 1.6x10^-8\n\n[Pb+2] = 1.6x10^-8/[SO4-2] = (1.6x10^-8)/(5.29x10^-4)\n[Pb+2] = 3.02x10^-5\n\nand there we have your answers:\n[Sr+2] = [SO4-2] = 5.29x10^-4 moles/liter\n[Pb+2] = 3.02x10^-5 moles/liter
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.
313595
Water... its got a pH=7
70269
Cu+S+O*4+H*2+O\n63.55+32.07+16*4+1.008*2+16= 177.64 g/mol\nThat x (usually a dot) means that there is a water molecule attached to the CuSO4.  The water molecule's mass should be included in the molar mass calculation.
903130
Na2SO4(aq)+ 2 AgNO3(aq)-->Ag2SO4(s)+ 2 NaNO3(aq).
351653
About 2000 mL, if each cup is 250 mL
1171925
PH of the aqueous solution of NaCl is 6 - 7 @ 1% by weight
1060616
Water H-O-H 100 °C 80 0.998  g/ml
249474
The mass of 1 cm^3 of water is 1 gramme.  Therefore, 100 cm^3 of water will have a mass of 100 g. If you add 150 g of sugar (or salt or anything else that does not produce an explosion or gives off a gas) the total mass will be 250 g.\n\nIf you had asked what is the total volume, that would have been a completely different situation.
24810
Cu + 2H2SO4 --> 2H2O + SO2 + CuSO4\nDouble the H2SO4 and the H2O.
472386
(250 X 0.35) + (1350X0.75) + (500X0.15) / (250+1350+500) = \n87.5+1012.5+75/(2100)\n1175 m mol/2100\n1.175 MOLES / 2100ml = 115g of sulfuric acid\n\n1.175moles /2.1 L = 0.58 M SOLUTION
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.
84557
A simpler formula will be SO2+NO2+CO2+H20 (unlimited)-> H2SO4 (strong and oxidant) +  HNO3 (nitric acid, stronger and oxidant) +  H2C03 (carbonic acid, weak but ´till corrosive, just like the soft drinks)