This will be classified as light on the API scale due to the large percentage of lighter fractions such as paraffins and naphthenes.
Answer:
a) 88.48%
b) 0.05625 mol
Explanation:
2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g) Reaction 1
CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g) Reaction 2
a) CH₃CH₂OH = 46.0684 g/mol
CH₃CH₂OCH₂CH₃ = 74.12 g/mol
1 mol CH₃CH₂OH ______ 46.0684 g
x ______ 50.0 g
x = 1.085 mol CH₃CH₂OH
1 mol CH₃CH₂OCH₂CH₃ ______ 74.12 g g
y ______ 35.9 g
y = 0.48 mol CH₃CH₂OCH₂CH₃
100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃
w _____ 0.48 mol CH₃CH₂OCH₂CH₃
w = 88.48%
b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.
Atoms of the same element with different mass numbers are called isotopes
<h3><u> Answer</u>;</h3>
= 4.0 L
<h3><u>Explanation;</u></h3>
Boyle's law states that the volume of a fixed mass of a gas is inversely proportional to pressure at a constant temperature.
Therefore; <em>Volume α 1/pressure</em>
<em>Mathematically; V α 1/P</em>
<em>V = kP, where k is a constant;</em>
<em>P1V1 = P2V2</em>
<em>V1 = 0.5 l, P1 =203 kPa, P2 = 25.4 kPa</em>
<em>V2 = (0.5 × 203 )/25.4 </em>
<em> = 3.996 </em>
<em> ≈ </em><em><u>4.0 L</u></em>
Answer:
You can mix both with rubbing alcohol, and rubbing alcohol is ionic.
Explanation:
To mix them together:
Combine 1 cup rubbing alcohol, 1 cup water, and 1 tablespoon of white vinegar into a spray bottle. The solution can kill germs and clean counters, according to Reader's Digest.
Rubbing alcohol is both polar and iconic.
Hope that this helps you and have a great day :)
