Answer:
<h3>1. B</h3><h3>2. A</h3><h3>3. B</h3><h3>4. B</h3><h3>5. C</h3><h3>I HOPE IT HELPS :) 100% sureness</h3>
Answer:
The answer to your question is 25.2 g of acetic acid.
Explanation:
Data
[Acetic acid] = 0.839 M
Volume = 0.5 L
Molecular weight = 60.05 g/mol
Process
1.- Calculate the number of moles of acetic acid
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = (0.839)(0.5)
-Result
moles = 0.4195
2.- Calculate the mass of acetic acid using proportions and cross multiplications
60.05 g ----------------------- 1 mol
x ----------------------- 0.4195 moles
x = (0.4195 x 60.05) / 1
x = 25.19 g
3.- Conclusion
25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M
<span>1 mole glucose gives 2 moles of ethanol
moles of glucose in 2.4 kg = 2400 / 180.18 = 13.320 moles
so moles of ethanol produced = 2* 13.32 = 26.64 moles
weight of ethanol 26.64 * 46.07
=1227.30 gm or 1.23 Kg</span>
