<span>An antifreeze is an additive which lowers the freezing point of a water-based liquid. An antifreeze mixture isused to achieve freezing-point depression for cold environments and also achieves boiling-point elevation to allow higher coolant temperature.</span>
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Answer:
A small still is separating propane and butane at 135 °C, and initially contains 10 kg moles of a mixture whose composition is x = 0.3 (x = mole fraction butane). Additional mixture (x = 0.3) is fed at the rate of 5 kg mole/hr. The total volume of the liquid in the still is constant, and the concentration of the vapor from the still (xp) is related to x, as follows: Xp = How long will it take for X, to change from 0.3 to 0.35.
Since there is no weight, I would assume that this is a 100g of pure compound.
Okay so I would be changing the percentage to gram to solve for the mole.
So
40.0g C (1 mol C/12.01 g C) = 3.33 mol C
6.73g H (1 mol H/1.01 g H ) = 6.66 mol H
53.3g O (1 mol O/16.00 g O) = 3.33 mol O
With that, two of our moles is 3.33, so we consider that are our 1, as it is also the lowest. Therefore the empirical formula is CH2O
