The answer would for sure be 178 because acid & base make that amount so yea!! I’m talllyyy right
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:
89.1 N
Explanation:
W = m gm = 55.0 x 1.62 = 89.1 N
The answer is (2) KNO3. This depends on the solubility of these four compounds at 10℃. For NaCl, it is 35.8 g, For NaNO3, 80.8 g. KCl, 31.2 g. KNO3, 21.9g. So only KNO3 is less than 25.0 g.
The answer is 98ppm.
The ppm (Parts per million) is also a concentration unit. 1 ppm is equivalent to 1mg/L
Now, concentration of the solution of sulfuric acid is 0.0980 g/L.
We rewrite, 0.0980 g = 0.0980*1000 mg = 98mg
Therefore, the concentration of the sulfuric acid solution is 98 mg/L = 98 ppm.
