Asked by Brianna
Three volatile compounds X,Y, and Z each contain elements Q. The precent by weight of element Q in each compound was determined. Some of the data obtained are given below
Percent by weight Molecular
Compound of Element Q Weight
X 64.8% ?
Y 73.0% 104.
Z 59.3% 64.0
(a) the vapor density of compound X at 27 degress C and 750.mm Hg was determined to be 3.53 grams per litre. Calculate the molecular weight of compound X.
(b) Determine the mass of the element Q contained in 1.00 mole of each of the three compounds.
(c) Calculate the most probable value of the atomic weight of element Q
(d) Compound Z contains carbon, hydrogen, and element Q. When 1.00 gram of compound Z is oxidized and all of the carbon and hydrogen are converted to oxidized, 1.37 grams of CO2 and 0.281 gram of water are produced. determine the most prbable molecular formula of compound Z
Hey, this is what I found on another website, actually. Part A seems to be fine, but I can't fathom how they got 64.2 as 73 percent of 104 in Part B. Part C is pretty confusing to me as well; hope you can make something of it:
a) For x; MM = dRT/P = 88
b) m = for x = 88. x 0.648 = 57 g
m = for y = 104 x 0.730 = 64.2g
m = for z = 64 x 0.593 = 37.95
c) % = M x 100/aw
M = (aw x%)/100
x = 37.95
y = 75.9
z = 37.95
Atomic weight is probably 37.95 ~= 38
Percent by weight Molecular
Compound of Element Q Weight
X 64.8% ?
Y 73.0% 104.
Z 59.3% 64.0
(a) the vapor density of compound X at 27 degress C and 750.mm Hg was determined to be 3.53 grams per litre. Calculate the molecular weight of compound X.
(b) Determine the mass of the element Q contained in 1.00 mole of each of the three compounds.
(c) Calculate the most probable value of the atomic weight of element Q
(d) Compound Z contains carbon, hydrogen, and element Q. When 1.00 gram of compound Z is oxidized and all of the carbon and hydrogen are converted to oxidized, 1.37 grams of CO2 and 0.281 gram of water are produced. determine the most prbable molecular formula of compound Z
Hey, this is what I found on another website, actually. Part A seems to be fine, but I can't fathom how they got 64.2 as 73 percent of 104 in Part B. Part C is pretty confusing to me as well; hope you can make something of it:
a) For x; MM = dRT/P = 88
b) m = for x = 88. x 0.648 = 57 g
m = for y = 104 x 0.730 = 64.2g
m = for z = 64 x 0.593 = 37.95
c) % = M x 100/aw
M = (aw x%)/100
x = 37.95
y = 75.9
z = 37.95
Atomic weight is probably 37.95 ~= 38
Answers
Answered by
Anonymous
i got 75.92 as 73% of 104. i have the same values for x and z though. i found Q to have an atomic mass of 19.
Answered by
Anonymous
this is stupid
Answered by
Anonymousss
I know right.
Answered by
anonymous
how the heck did this person get .730*104 to equal that? it should definitely be 75.92. also his/her logic for c makes no sense...
Answered by
Anonymous
This has a few incorrect calculations. Also incorrect copying. Please do not use these answers.
Answered by
Anonymous
go back to school and relearn math
Answered by
Anonymous
what about part d
Answered by
Anonymous
a.) THE MOLAR MASS IS 88
c.) The mass of the element is 19
d.) C2H2F2
c.) The mass of the element is 19
d.) C2H2F2
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