Answer:
<h2>━☆゚.*・。゚Fine, but not as good</h2>
Answer:
-0.1767°C (Option A)
Explanation:
Let's apply the colligative property of freezing point depression.
ΔT = Kf . m. i
i = Van't Hoff factot (number of ions dissolved). Glucose is non electrolytic so i = 1
m = molality (mol of solute / 1kg of solvent)
We have this data → 0.095 m
Kf is the freezing-point-depression constantm 1.86 °C/m, for water
ΔT = T° frezzing pure solvent - T° freezing solution
(0° - T° freezing solution) = 1.86 °C/m . 0.095 m . 1
T° freezing solution = - 1.86 °C/m . 0.095 m . 1 → -0.1767°C
Answer: a) 
acid : hydronium ion
base : methoxide ion
conjugate acid : methanol
conjugate base: water
b) 
acid : hydrogen chloride
base : ethoxide ion
conjugate acid : ethanol
conjugate base: chloride ion
c) 
acid : methanol
base : amide ion
conjugate acid : ammonia
conjugate base: methoxide ion
Explanation:
According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.
The species accepting a proton is considered as a base and after accepting a proton, it forms a conjugate acid.
The species losing a proton is considered as an acid and after loosing a proton, it forms a conjugate base
For the given chemical equation:
a) 
acid : hydronium ion
base : methoxide ion
conjugate acid : methanol
conjugate base: water
b) 
acid : hydrogen chloride
base : ethoxide ion
conjugate acid : ethanol
conjugate base: chloride ion
c) 
acid : methanol
base : amide ion
conjugate acid : ammonia
conjugate base: methoxide ion
.