Answer:
put a filter to see if there is any stagering solution
Explanation:
There are several ways to visually represent compounds. For this particular organic compound, we can use the skeletal formula and the expanded formula. The skeletal makes use of lines to show which atoms are bonded to each other. The expanded formula shows the species of the atoms and their bonding with other atoms. I have attached the two representations.
I dont know your question but that is true
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:
2.29 g of N2
Explanation:
We have to start with the <u>chemical reaction</u>:

The next step is to <u>balance the reaction</u>:

We can continue with the <u>mol calculation</u> using the molar mass of
(65 g/mol), so:

Now, with the<u> molar ratio</u> between
and
we can <u>calculate the moles</u> of
(2:3), so:
With the molar mass of
we can <u>calculate the grams</u>:
I hope it helps!