the question was too long for this section so it is just down below any help is appreciated ?
A 6.05 g sample of water is introduced into a 0.838 L flask containing some C2H2 gas. The flask is heated to 215.77 °C at which temperature all of the water is converted to the gaseous phase, giving a total pressure in the flask of 19.659 atm.
Calculate PH2O (in atm) in the flask at 215.77 °C. Report your answer to three decimal places in standard notation (i.e. 1.234 atm).
- Roger the MoleLv 71 month agoFavourite answer
Ignore the C2H2.
P = nRT / V = ((6.05 g H2O) / (18.01532 g H2O/mol)) x (0.082057366 L atm/K mol) x
(215.77 + 273.15) K / (0.838 L) = 16.078 atm H2O
[There's the requested "three decimal places", although only one is justified by the rules about significant digits.]
- davidLv 71 month ago
A 6.05 g sample of water is introduced into a 0.838 L flask
6.05 / 18.0 = 0.336 mol H2O
PV = nRT
P = nRT/V = 0.336(0.821)(1488.77)/0.838
P = 490.079 atm