250mL of 0.35 M H2SO4; 1.35 of 0.75 M H2SO4; 500mL of 0.15 M H2SO4 are mixed. How many g of H2SO4 in mixture?
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
590452
(NH4)2 SO4  + (FeSO4)2 + 2H2O____(\nNH4)2SO4Fe2(SO4)3.24H2O\n\nAmmonium ferric sulphate
220540
Here's a good chart to use\nhttp://www.simetric.co.uk/si_medical.htm\nTo answer your question there are 1000 mcg to 1 mg so 2.5 mg equals 2500 mcg
1012209
300mL\n    x.18\n_______\n\n   54.00 mL
951445
... what's the specific gravity of the mixture?\nSince the total volume is the sum of the two ingredients, there was no volume loss due to mixing. (Sometimes there is; water and ethanol are an example.) Then the total mass is 6*0.8 + 5 * 1 = 9.8 kg and the sp. gr. = 9.8/11 = 0.8909 kg/L.
24810
Cu + 2H2SO4 --> 2H2O + SO2 + CuSO4\nDouble the H2SO4 and the H2O.
726441
4.74 / 300 = 0.0158 g/ml\nin 35 ml\n0.0158 * 35 = 0.553 g\nDividing by 40 g/mol (MW) we get\n0.553 / 40 = 0.0138 mol in 100 ml\nSo in 1000 ml the number of moles will be 0.138\nThe molarity is then 0.138 M.
787481
NaHCO2 + H2SO4 --> Na2SO4 + CO2(g) + H2O(L)\n\nSodium bicarbonate + sulphuric acid equals sodium sulfate + carbon dioxide + water
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
748321
You're kidding right? \n\n1L (1000ml/ 1L) x (0.5g/1ml) = 500g
938236
Step one: find the moles of H2SO4.  (2.7 moles per liter)*0.025 liters.\n\nStep two: multiply by two because the reaction is \n2NaOH + H2SO4 ----> 2H2O  + Na2SO4 and so you need two NaOH molecules per molecule of H2SO4.\n\nStep three: multiply by the molecular mass of NaOH.\n\nThat will get you the answer.
854886
1. 1,000  (each full number in the pOH scale is a factor of ten in the concentration).\n\n2.  C, because you have about 98 grams of H2SO4\n\n3.  b is an acid and its salt.\n\n4.  d because you have about five moles of water.
969093
H2 = 2g/mol   2.05g = 1.025mol\nCO = 28g/mol    15.85g = .57mol\n\n2H2 + CO --> CH3OH\n\nmust be reduced to \nH2 + (1/2)CO --> (1/2)CH3OH\n\n1.025mol + (1/2).57mol --> the quantity of the limiting reactant which is .285mol CH3OH which at 32g/mol = 9.12g
116087
keeping with the same unit fractor harry is correct.\n \n .2 ml
905904
step 1 2NaOH + H2SO4 ---------> Na2SO4 + 2H2O\nstep 2 amount of NaOH = 0.2 x 25/1000= 5.0 x 10^-3 mol\nstep 3 amount of H2SO4 = s.0x10^-3 x 1/2 = 2.5 x 10^-3 mol\nstep 4 mass of H2SO4 = 2.5 x 10^-3 x 98 = 0.245g required\n\nNOTE\n^= power\nin step 4 it is multiplied by 98 because that is the molar mass of H2SO4 ie (2x1)+32+ (16 x 4) =98\n\ni do hope this is correct it has a been about 3 yrs since i did my A Level in chemistry
502184
5H2O2(aq)+2KMnO4(aq)+\n3H2SO4(aq)-->\n2MnSO4(aq)+K2SO4(aq)+\n8H2O(l)+5O2(g)
124546
concentrated sulfuric acid is 36N, so if you dilute it by 17.9 times(56mls to 1L) then it would be 36 divided by 18(rounded off) or 2N.