A sample of hydrogen gas was collected over water at 21 Degrees C and 685 mmHg.  The Volume of the container..
926935
PV = nRT\n\n685 mmHg / 760 = .901 atmospheres\n\n18.6 mmHg / 760 = .0245 atm\n\n.901-.0245 = .8765 atm\n\n21 degrees c = 294 K\n\n.8765 atm x 7.80 L = n x .08206 x 294 K\n\nsolve for n then use the molar mass of H2 (2.016) to convert to grams.\n\nmy calculator's out in the car or else i'd do it for you.
1143445
Water has a density maxima at 4C
706503
find the heat evolved, assuming heat cap of water is 4.2J/g/K\n\n(31.43-25.32) x 4.2 x 100g = 2566.2J\n\nnow according to definition, molar heat of neutralisation is calculated upon the mols of water formed. this number of mols is equivalent to the no. of mols of the limiting reagent (in this case, the HCl).\n\nmol of HCl= 0.050 x 1 = 0.05mol\n\nenthalpy of neutralisation= 2566.2/0.05= 51324J/mol\n\nthis answer is pretty close to real-life experimental values for strong acid-strong base.
643686
It would make water (H2O).
367911
For example, hydrogen burns in oxygen to form water vapor. \n\n2H2(g) + O2(g) -> H2O(g) \nHow much hydrogen must be burned to form 1.00 L of water vapor at 1.00 atm pressure and 3000C? \nTo work this problem, we need to figure out how many moles of water vapor we need to form. Use the ideal gas equation to compute this: \n\nPV = nRT \n1.00 atm * 1.00 L = n*0.0821 (L*atm/mol*K)* 473 K \nn = 0.0258 moles \nThus, we need to form 0.0258 moles of water. Now it's just a basic stoichiometry problem: for each mole of water formed, you need 2 moles of hydrogen gas, so 0.0258 moles water * 2 moles H2/1 mole water = 0.0516 moles of hydrogen \n\n=====================================================\n\nthis will help you a lot\n\nhttp://www.chem.hawaii.edu/asato161/Ch10NotesP2.pdf#search='Gas%20Stoichiometry%20PV%3DnRT'\n\nhttp://members.aol.com/profchm/gasstoic.html
480599
Hydrogen gas was generated on site by the action of sulphuric acid on iron filings in a large, covered container.  The gas generated was then pumped into the balloon.
875948
Water brakes down into h and o. 2H2O.....2H2 +O2.
192134
1.74g should = 17.4g approx.\n\nso,\n\n x  = H (Hydrogen) = 1.008g
693067
Water, nitrogen and carbon.
970781
Potential of Hydrogen.
247135
The formula for water is H2O i.e. there are 2 atoms of hydrogen and one atom of oxygen. Its molecular weight is 18 g/mol .
56363
Dalton's law states that the total pressure is equal to the sum of the individual partial pressures of the gases in the mixture: therefore if the vapor pressure of water @ 21.5 degrees C is 2.6kPa and the total pressure in the flask is 99.4kPa then your answer should show 99.4-2.6=96.8kPa\n\nHope that helps...
930579
24 litres of air (at RTP) will contain 1 mole of gas. You are told that you have 12.9 cubic kilometres of water vapour. Convert this to litres, and then to moles The enthalpy of vaporisation of water is 40.65kJ/mol,  so you can now work out the total energy released.\nYou would also need to research the power consumption of a city of 1 million people!
870041
This question does indeed deal with Dalton's law with the stipulation that the total pressure of the "wet gas" is referring to the oxygen collected in gas form combined with the water vapor collected from water evaporation.  Therefore, all you need to do is figure out what the pressure of water vapor is at 20 degrees Celsius.  Convert to mm Hg if you need to and there you have it.  Subtract the two and what's left is the contribution of the oxygen.  \n\n760 mm Hg = 101.325 kPA just for quick reference.  Hope this helps!
984937
The equation is: 2H2 + O2 ---> 2H2O\nSo you would form 2L of water VAPOR\nIf you want this in terms of LIQUID water, remember that one mol of a gas occupies 22.4L, so you have 2/22.4 mols of water (MW=18) formed, so...\n2/22.4 x 18 = 1.6 grams or 0.0016L of LIQUID water.\nThis assumes that all three gasses act as 'ideal gasses' (which is only close) and conditions are reduced to STP...so answer is not exactly correct, but 'close enough for government work'  :-))
837250
The equation doesn't balance.  There's no hydrogen on the right side.  But, assuming that you ended up with H2 gas on the right (I don't think that reaction works real-world, but whatever), the main thing you need to look at is the chlorine.  It's Cl2 on one side and Cl3 on the other.  That means that you must have a total number of chlorine atoms which is a multiple of 6.  Try balancing the equation with 6 chlorine atoms and see what you get:\n3CuCl2 + 3H2O + 2Al -----> 2AlCl3 + 3 CuO + 3H2
852032
try wikipedia.org, it can tell you anything.... body is mostly water, I believe.  More than 50% H20