albertlee Posted November 12, 2004 Posted November 12, 2004 A gas X is 42g/mol A gas Y is 56g/mol If both gas are in mixture, the mixture is 42g/mol which gas in that mixture is more abundant??? X or Y?? Can any one explain to me in details as well?? thx Albert
YT2095 Posted November 12, 2004 Posted November 12, 2004 gas X, as any introduction of gas Y would increase the value. there`s no mix, it`s a trick question
albertlee Posted November 12, 2004 Author Posted November 12, 2004 I know they dont actually chemically "mix" together..... Can you give me a bit of chemical and mathematical manipulation?? Albert
albertlee Posted November 12, 2004 Author Posted November 12, 2004 Sorry, for duplicating post... but could any one help me?? I am stuck on this.... thx in advance !!
YT2095 Posted November 13, 2004 Posted November 13, 2004 x=42 y=56 the "Mix" can`t be if they are the only 2 gasses. 42 is the least it can be if only X is used, ANY introduction of Y would take that value higher! it would be different if there were a 3`rd gas lower than 42 or gas X was lower than 42, but there isn`t. there`s no mix!
albertlee Posted November 13, 2004 Author Posted November 13, 2004 Ohhhhhhh so.... when the two gases are placed together, their g/mol is 42 in average right?? In average, both sides need to offer 42g....., the higher g/mol must give less amount of moles but this question is not actually correct to me. There is 4/7 of gas X and 3/7 of gas Y, according to my calculation...., and if 4/7 is 1 mol, then 3/7 should be less....., then when we divide 84g (the total mass) by the amount of the total mols, the result couldn't be 42g/mol Any help? Albert
YT2095 Posted November 13, 2004 Posted November 13, 2004 when the 2 gasses are mixed its (42+56)/2 to give the G/Mol. and that`s 49 g/mol. any introduction af gas Y to gas X will always be higher than 42, no matter how small an amount.
Primarygun Posted November 13, 2004 Posted November 13, 2004 In a equation, for two different types of composition, Let z% be the composition of x and (100-z)% be the composition of y. The new weight=[xz+y(100-z)]/100
albertlee Posted November 13, 2004 Author Posted November 13, 2004 So, YT, do you mean that there is even none of gas Y in the so-called mixture?? Albert
YT2095 Posted November 13, 2004 Posted November 13, 2004 that`s exactly what I mean. based upon the parameters in your original post yes
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