Answer: Too much base was added
i guessed
Explanation:
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.
B. a circle graph
circle graphs are the best to show percentages because they’re very easy to look at and get info from
Answer: this question is 3 days ago? Omg
Answer:
The required volume of hexane is 0.66245 Liters.
Explanation:
Volume of octane = v=1.0 L=
Density of octane= d = 
Mass of octane ,m= 
Moles of octane =
Mole percentage of Hexane = 45%
Mole percentage of octane = 100% - 45% = 55%
Total moles = 11.212 mol
Moles of hexane :
Moles of hexane = 5.0454 mol
Mass of 5.0454 moles of hexane,M = 5.0454 mol × 86 g/mol=433.9044 g
Density of the hexane,D = 
Volume of hexane = V
(1 cm^3= 0.001 L)
The required volume of hexane is 0.66245 Liters.
