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natulia [17]
3 years ago
13

I have to obtain an empirical formula from:

Chemistry
1 answer:
JulsSmile [24]3 years ago
7 0
When finding empirical formulas decimals are fine so you wouldn't have to multiply all of them to turn into whole numbers. What you would do is divide each of those numbers by the smallest of those numbers and take the nearest whole number so if you had:

1.81 mol     .6 mol     and .3.61 mol you would divide each of those numbers by .6
and you would have 3, 1, and 6

for your problem you would divide them all by 1 and Carbon would be 2 (if your teacher normally has you round down at .5's then it would be 1)  hydrogen would be 3 and oxygen would be 1. 
C2H3O

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A gas has a volume of 590 mL at temperature of -55.0 C. What volume will the gas occupy at 30.0 C show your work
DENIUS [597]
Data:
V_{initial} = 590\:mL
T_{initial} = -55.0^0C
converting to Kelvin
TK = TC + 273
TK = -55.0 + 273 → TK = 218.0 → T_{initial} = 218.0\:K
V_{final} = ? (in\:milliliters)
T_{final} = 30.0^0C
TK = TC + 273
TK = 30.0 + 273 → TK = 303.0 → T_{final} = 303.0\:K

By the first Law of Charles and Gay-Lussac, we have: 
\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }

Solving:
\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }
\frac{ 590 }{ 218.0 } = \frac{ V_{f} }{ 303.0 }
Product of extremes equals product of means:
218.0* V_{f} = 590*303.0
218.0 V_{f} = 178770
V_{f} = \frac{178770}{218.0}
\boxed{\boxed{V_{f} \approx 820.04\:mL}}\end{array}}\qquad\quad\checkmark
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xz_007 [3.2K]

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