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.
Answer:
87.5 mi/hr
Explanation:
Because a = Δv / Δt (a = vf - vi/ Δt), we need to find the acceleration first to know the change in velocity so we can determine the final velocity.
vf = 60 mi/hr
vi = 0 mi/hr
Δt = 8 secs
a = vf - vi/ Δt
= 60 mi/hr - 0 mi/hr/ 8 secs
= 60 mi/hr / 8 secs
= 7.5 mi/hr^2
Now that we know the acceleration of the car is 7. 5 mi/hr^2, we can substitute it in the acceleration formula to find the final velocity when the initial velocity is 50 mi/hr after 5 secs.
vi = 50 mi/ hr
Δt = 5 secs
a = 7.5 mi/ hr^2
a = vf - vi/ Δt
7.5 = vf - 50 mi/hr / 5 secs
37.5 = vf - 50
87.5 mi/ hr = vf
