Answer:
This is due to more hydrogen bonding in ethylene glycol than it is in isopropyl alcohol
Explanation:
The boiling point of isopropyl alcohol is 82.4 °C it contains only a single OH group, hence intermolecular hydrogen bonding is solely responsible for it's boiling point, whereas Ethylene glycol (CH2OHCH2OH) contains 2-OH group and both intermolecular and intramolecular hydrogen bonding are responsible for the higher boiling point of ethylene glycol at 198 °C.
Answer:
The answer to your question: 0.7 M
Explanation:
Data
V of KOH = 90 ml
[KOH] = ?
V H2SO4 = 21.2 ml
[H2SO4] = 1.5 M
2KOH(aq) + H₂SO₄(aq) → K₂SO₄(aq) + 2H₂O(l)
Molarity = moles / volume
moles of H₂SO₄ = (1.5) (21.2)
= 31.8
2 moles of KOH -------------- 1 mol of H₂SO₄
x -------------- 31.8 mol of H₂SO₄
x = (31.8)(2) / 1
x = 63.8 moles of KOH
Molarity = 63.8 / 90
= 0.7 M
Answer:
Explanation:
Hello!
In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:
Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:
Best regards!
You can tell the difference by How shiny it is and by how heavy it is
