correct me if I am wrong, but I got .12 mol of CO2. I agree with your method up until Pco2= (Pt) nco2/nt. then I differ from yours because I solve for the mol fraction and put Xco2 equal to nco2/nt. I use the value calculated for x and the given value for nt to get .12mol. I could be wrong though....we'll see if someone else weighs in....

I agree with Sahil, I also got .12 mol of CO2, but we could be wrong as well. I follwed the same process as Sahil so you might want to go back and check and see if you agree. Other than that great job! I definitly agree with the first few steps of your process!

Good job on your blog. I liked how you used the program to get the subskript, that helped. When I worked it I got .122 moles so either my sig figs or ytour sig figs are wrong.

correct me if I am wrong, but I got .12 mol of CO2. I agree with your method up until Pco2= (Pt) nco2/nt. then I differ from yours because I solve for the mol fraction and put Xco2 equal to nco2/nt. I use the value calculated for x and the given value for nt to get .12mol. I could be wrong though....we'll see if someone else weighs in....

ReplyDeletealso for the partial pressure, when you calculated Pco2 I think you were off... 540+320=860, not 840. I think Pco2 should be 520 torr

ReplyDeleteI agree with Sahil, I also got .12 mol of CO2, but we could be wrong as well. I follwed the same process as Sahil so you might want to go back and check and see if you agree. Other than that great job! I definitly agree with the first few steps of your process!

ReplyDeleteGood job on your blog. I liked how you used the program to get the subskript, that helped. When I worked it I got .122 moles so either my sig figs or ytour sig figs are wrong.

ReplyDelete