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The amount, in grams, of ethanol needed to prepare 268 grams of a 9.76 M solution of ethanol in water will be 120.50 grams
<h3>Solution preparation</h3>
In order to prepare 268 grams of solution, 268 mL of water would be needed because 1 mL of water weighs 1 gram.
Since we now know the volume of solution that we want to prepare, the amount of solute (ethanol) that would be required in order to make a solution of 9.76 M will be:
Mole required = 9.76 x 268/1000 = 2.62 moles
Mass of 2.62 moles ethanol = 2.62 x 46.07 = 120.50 grams
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Answer:
106.95 lb of CO2
Explanation:
Equation of reaction:
2C8H18 + 25O2 = 8CO2 + 18H2O
Mass of C8H18 = density × volume = 0.692×12×3.785×1000 = 31430.64 g
From the equation of reaction above,
2 moles of C8H18 (228 g) produced 8 moles of CO2 (352 g)
31430.64 g of C8H18 will produce 31430.6×352/228 = 48524.50 g of CO2 = 48524.50/1000 = 48.52450 kg of CO2 = 48.52450×2.204 = 106.95 lb of CO2
Answer:
The density.
Explanation:
The density is an intensive property of fluids, defined as mass divided by volume.
This property is directly related to the intermolecular forces that are present in the substance: the stronger the intermolecular forces, the higher the density will be, and viceversa. Therefore, organic fluids usually have lower densities than water, because <u>the intermolecular forces in organic fluids are not as strong as in the intermolecular forcer in water</u> (water has very stable hydrogen bonds, whereas organic fluids have van der Waals, weaker than hydrogen bonds).
The liquid that has the lower density will be the the liquid that is on the top layer. In this case, the organic compound tert-butyl methyl ether will be on top of water.