Recent content by hclark91

sorry typo PV=nRT 2n=PV/RT=(0.66atm)*(1L)/(0.0821 L-atm/mol-k)*(298K)=0.0269 mol n=0.0135 mol 0.0135 mol * 51g/1mol NH4HS= 0.6879 g NH4HS .1 mol *51g/1 mol= 5.1 g 0.6879g/5.1g=0.135 0.135*100=13.5% dissociated I accidentyl put an extra 0 in 0.06879 it should be 0.6879. The rest of the numbers...

Using the method I thought would work I got PV=nRT (0.66atm)(2L)=n(0.0821L-atm/mol-k)(298k) n=0.5395 0.5395x51g/1mol= 2.752g of NH2HS to obtain equilibrium does this seem right??

Homework Statement I am having trouble with this question can anyone point me in the right direction? Solid NH4HS dissociates according to the equation: NH4HS (s) ---> NH3 (g) + H2S (g) SOme solid NH4HS is placed into an initially evacuated vessel and when equilibrium is established there is...