Answer: B: Your ears send a message through the nerves to the brain.
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.
Combustion is a reaction between a combustible substance and oxygen, to ultimately produce carbon dioxide and water. Reaction between carbon and oxygen would give,
C + O2 ------> CO2
Here, we have 86.5 grams of carbon dioxide, CO2, which is a product of combustion. Dividing this mass by the molar mass of CO2, which is 44 grams, we can determine the number of moles of CO2.
<u> 86.5 g CO </u> = 1.966 moles CO2
44 g CO2/ mole
Considering that CO2 is composed of 1 mole of carbon and 2 moles of oxygen, and that with complete combustion, 1 mole of carbon reacts to produces 1 mole of CO2, we can then determine the mass of the carbon in the hydrocarbon fuel.
1.966 moles CO2 x <u> 1 mole C </u> x <u> </u><u>12 g C </u> = 23.59 g C
1 mole CO2 1 mole C
We were given 25.0 grams of the fuel hydrocarbon. A hydrocarbon is a substance consisting of carbon and hydrogen. To determine the mass of the hydrogen in the fuel, we simply subtract 23.59 grams from 25.0 grams.
25.0 g - 23.59 g = 1.41 grams Hydrogen
To know the number of moles of hydrogen, we divide the mass of the hydrogen in the fuel by the molar mass of hydrogen, which is 1.01 g/mole. Thus, we have 1.396 mole hydrogen.
To determine the empirical formula, we divide the number of moles carbon by the number of moles hydrogen, and find a factor that would give whole number ratios for the carbon and hydrogen in the fuel,
Carbon: <u> 1.966 mol </u> = 1.408 x 5 (factor) = 7
1.396 mol
Hydrogen: <u> 1.396 mol </u> = 1.00 x 5 (factor) = 5
1.396 mol
Thus, the empirical formula is C7H5
Elector would be that a laser converts electrical energy