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:
False
Explanation:
The density of Interstellar dust is very low,yet it still blocks starlight because. .....the Dust emission nebulae like M42 occur only near star that emit large amount of. ......Hydrogen is the major gas in the interstellar medium.
I think is 5 because you use density is found by mass divide by volume.
Answer:
The answer is Kr (Krypton).
This is because krypton has an electronic configuration of:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6
Taking note of the sequence of electronic configuration:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s
It can be seen that Kyrpton's electronic configuration finishes just before the 5s subshell. Therefore, the noble gas notation for an element with valence electrons in the 5s subshell can use [Kr] as a shortcut to denote its electronic configuration. For example:
If an element has 1 valence electron in the 5s subshell, the noble gas notation will be:
[Kr] 5s1
Explanation:
The first one is the correct answer: <span>The potential energy of the products is greater than that of the reactants and the change in enthalpy is positive.<span> </span></span>
