what is the percent of water in MnSO4.7H2O?
371632
First find the mass of 1 mole of MnSO4.7H2O Then find the mass of 7 moles of H2O. Now find the % by dividing the latter by the former, then multiply by 100. I get about 45%.\n\nEriochrome Black T at pH 10 is blue (in the absence of cations such as Mg2+ and Ca2+).
709905
There are two acidic protons on H2SO4, so normality is 2*molarity.\nanswer is 0.3 N
787311
A. H2O. Molecular water is 2H2O
867817
25cm3 of a 4M solution of CuSO4 contains 25/1000 x 4 M CuSO4 = 0.1M\n\n! mole of CuSO4 weighs 160g (Cu = 64 + S=32  + O4 = 64)\n\nSo   0.1 Mole weighs  16g. This is how much anhydrous solid is needed. \nHowever copper sulphate usually comes as crystals and 1 mole of this weighs 250g ( + 5H20 = 90) so to make a 0.1 molar solution from crystals you need 25g CuSO4. 5H2O
216402
It's:\n\nBaSO4(s)---water <--> Ba2+(aq)  +  SO4 2-(aq)\n\nthe solubility for this compound is very low  per  g/L\n\n\n about 0.0023g/L
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.
438725
Magnesium Sulfate\n\nMgSO4\n\nMg(OH)2 + H2SO4 -> MgSO4 + 2HOH\n\nHOH = H20\nJust in case...
751490
Without doing a rigorous calculation I will guesstimate that it will be Oxygen.\n\nIf body is 70% water, Oxygen in water is ~90% so oxygen will be ~63% in body.
915044
7.04 grams\n\n15 grams of water is .44 mol and according to the equation, the ratio between water and oxygen is2:1 so if .44 mol of water is decomposed. then .22 mol of oxygen will be produced. and .22 mol of oxygen is .22 *32 =7.04grams
288276
ρwater ~ 1g/ml\n\nq = 1000g x 4.18J/ºC.g x (37ºC-0ºC) = 155.660kJ (1.6x10^5J to correct Sig Figs)\n\nsecond part. The amount of heat required in part 1 will melt 155.660kJ / 6.01kJ/mole = 25.9 moles of ice or 468g
712851
Dissolve 3.2 g of KMnO4 in water to make 500 mL final volume. This will be 0.04M KMnO4 solution.
583365
First of all work out the molar mass of CuSO4*5H2O\n Cu + S + (9 * O) * (10 * H)\n\nSorry I don't have a chemical table next to me\n\nCall this A\n\nSecondary work out the mass of 5 moles of water.\nCall this B\n\n(5 * O) + (10 * H)\n\nthe theoretical % is then B/A * 100 = x%
975769
Al(s) + MnO4- (aq) -----> MnO2 (s) + Al (OH)4- (aq)\n\nSplit the above into two half equations:\n\n1) Al(s)  --->  Al (OH)4- (aq)  and\n2) MnO4- (aq) -----> MnO2 (s)\n\nBalance 1) as follows:\n\n1) 4OH^-  +  Al(s)  --->  Al (OH)4- (aq) now balance for charge:\n1) 4OH^-  +  Al(s)  --->  Al (OH)4- (aq)  +  3e^-\n\nBalance 2) as follows:\n2) MnO4- (aq) ---> MnO2 (s)  +  2H2O (l) (turn xs O into H2O)\nNow balance with H^+:\n2) 4H^+  +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)\nnow balance charge:\n2) 3e^-  +  4H^+  +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)\n\nNow add 1) and 2) (electrons happen to be equal):\n4OH^-  +  Al(s)  +  4H^+  +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)  Al (OH)4- (aq)\nnow OH^-  +  H^+  ---> H2O:\n\n4H2O  +  Al(s)    +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)  Al (OH)4- (aq)\n\nSubtract 2H2O from both sides:\n2H2O (l)  +  Al(s)    +  MnO4- (aq) ---> MnO2 (s)  +    Al (OH)4- (aq)\n\nI hope that helps. If you need more help, please email me or re post.
127716
based on the given balances equation.\n\nthe volumes could be:\n  \nSTP = 22.4L per mole\n\n 14.2L SO2 ( 1 mol H2S / 2 mol SO2) = 7.01 L of H2S\n\n and\n \n14.2L SO2 (3 mol O2 / 2 mol SO2 )  =  21.3 L of O2
780730
K2(C2O4)*H2O is Potassium oxalate monohydrate.  The oxalate is oxidized to CO2 (Carbon dioxide).\n\nKMNO4 is Potassium permanganate.  The permanganate is reduced to Mn+2 (Manganese II cation).  This requires an acid (H+).\n\n(C2O4)-2 (aq) + (MnO4)-1 (aq) + 8 H+ (aq) --> 2 CO2 (g) + Mn+2 (aq) + 4 H2O (l)
900769
Chemically it is MgSo4 or Magnesium Sulfate. I would ask for this or write it down and hope the other person has a minium education in chemistry ....
867788
Liquid.  Melting point is 600.65K