The solubility of hydrogen in water at S.T.P. is 0.00193 g L-1. At what pressure will 1 litre of?
403786
.2g/L ( 1atm / .00193g/L) =  103.63 atm
768648
Actually it is not a solubility problem, but a buffer.\n\nLet's use the Henderson-Hasselbalch equation for simplicity\n\npH=pKa+log[CH3COONa]/[CH3COOH]\n\nIf V is the volume of the solution in L then\n[CH3COOH]= mole/V= 0.010/V\n[CH3COONa]= mole/V =x/V\n\nso substitute in the equation and you get\n\n5.37=4.74+ log((x/V)/(0.01/V))=>\n0.63=log(x/0.01) =>\nx/0.01= 10^0.63 =>\nx =0.01*10^0.63= 0.0427 mole =42.7 mmole or since MW=82\nm= 0.0427*82= 3.50 g
598989
The amount of hydrogen by mass in water is 1/9.\nLiquid hydrogen has a density of roughly 1/14 that of water.\nThus the volume of liquid hydrogen would be 14/9 gallons, i.e. more by volume than you started with (Sounds wrong but it's true).\nIf you want me to convert it into terms of weight you'll have to use SI units. I always though a gallon was a warship they had back 400 years ago.
182906
pH is negative log of the concentration of hydrgen ions.\n\nSo pH of 1 would mean 10^(-1) or 0.1 concentration of hydrogen ion.\n\npH of 14 would mean 10^(-14) or 0.00000000000001 concentration of hydrogen ions.\n\nThat would mean that with a no. between 1 and 14 we can express a very wide range of concentrations.
597779
The first answer is poh=3.167 and the ph=10.832\nThe second answer is poh=2.119 and the poh=11.881\n\nNot sure onthe concentration of hydrogen though.
295517
First you need to know what a mole is. A mole of HCl is it's formula mass, which is 36.5g. So, if you put 36.5g of HCl into one litre of water, you would have 1.0 molar solution. For 0.1 molar you use 3.65g HCl.
853370
The concept of 1.5M hydrogen gas is rather odd, but, yes, you use the Nernst Equation.  Don't forget to convert the pH to a molarity of H+ ions.
737133
10,000 Litres of what?\n\n1,000 Litres of water is a metric ton or 1, 000 kg.\nSo if yoiu have 10,000 Litres of water it will weigh 10 metric tonnes or 10,000 kg.\n\nDifferent liquids, solids and gasses have different densities so 10,000 Litres of them will have different weights.
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
482843
Boyle's law\n\nP1xV1=P2xV2\n\n(.75 atm)x(2 L) = (2 atm)x(X)\n\n1.5=2x\n\ndivide both sides by 2\n\n.75 L would be the volume of helium if the container was at a pressure of 2 atm.
813197
0.0821 moleP1V1/T1=P2V2/T2\n375*4/293=760*V2/273.5\nV2=375*4*273.5/(293*760)\n22.4 litres contains 1mole\n375*4*273/293*760\n=375*4*273/293*760*22.4\n=0.0821 mole
192134
1.74g should = 17.4g approx.\n\nso,\n\n x  = H (Hydrogen) = 1.008g
928965
molarity  = moles / liters\n\nmoles = given mass / molecular mass\n\n10 g H2SO4 / 98 g per mole = about  0.1 moles\n\n0.1 moles / 1 L = 0.1 M solution
134712
This question is based on the gas law PV=nRT where \n\nP= pressure in atm (1atm=101.3kPa)\nV= volume in litres\nn= # of moles\nR= .08206\nT= temp. in Kelvin (273.15K=1*C)\n\nand the equation\n\nP1V1=P2V2
907196
The concentration of H+ is 3.0 x 10^-4, so you know that the pH will be around 4.  What you need is a calculator that has a LOG button to convert the 0.0003 to an integer value.\n\nSo on your calculator you press the LOG button and then type in 0.0003 and then a ")" and press ENTER.  \n\nYou should get a result that says: -3.522878745 or so.  You need to round off to -3.5 and since pH is the NEGATIVE log of the hydrogen ion concentration you get rid of the minus.\n\nThe answer is 3.5.\n\n1x10^-3.5 is approximately equal to 0.0003.
533630
water is incompresible, so ...\n\nVgas = Vtank - Vwater\n\nm(gas) = M(gas)*Niu(gas)\n\nNiu(gas) - no. of moles \nM(gas) - molecular mass of gas\n\nNiu(gas) = m(gas)/M(gas)\n\nwe have the following equation:\n\nPV = Niu*R*T, where:\n\nP - pressure (N/m^2)\nV - volume (m^3)\nNiu - no. of moles of gas\nR - universal gas constant (8.31 in these conditions)\nT - temperature (K)\n\nP = Niu(gas)*R*T/V(gas)\n\nP = [m(gas)/M(gas)]*R*T/[Vtank - [m(water)/(water density)]]\n\nin formula introduce Vtank in m^3 (not in liters 1 m^3 = 1000 dm^3 = 1000 liters), and the pressure will result in N/m^2\n\ntransform it in psia ... and the probleme is solved
1108940
Well, it doesn't, really, except that since the law of definite composition says that substances always have the same ratio of the same atoms, they're going to have the same properties, including density (mass per unit volume).  Water is always two hydrogen atoms to one oxygen atom, and will always have a density of 1.00 grams per milliliter at 4 degrees Celsius.  Same applies to all other substances.  In terms of it relating directly to volume, though, it doesn't.