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
Mass of H₂ needed to react with O₂ : 1.092 g
<h3>Further explanation</h3>
The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight / volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.
Reaction
O₂(g) + 2H₂(g) → 2H₂O(g)
mass of O₂ : 8.75 g
mol O₂(MW=32 g/mol) :

From the equation, mol ratio of O₂ : H₂ = 1 : 2, so mol H₂ :

Mass H₂ (MW=2 g/mol) :

Answer:
Boiling point for the solution is 100.237°C
Explanation:
We must apply colligative property of boiling point elevation
T° boiling solution - T° boiling pure solvent = Kb . m
m = molalilty (a given data)
Kb = Ebulloscopic constant (a given data)
We know that water boils at 100°C so let's replace the information in the formula.
T° boiling solution - 100°C = 0.512 °C/m . 0.464 m
T° boiliing solution = 0.512 °C/m . 0.464 m + 100°C → 100.237 °C
Answer:
A molecule is stable when there is no energetically-accessible mechanism available that allows it to reacts and form a more stable molecule or molecules whiles Are configuration of unknown atomic nuclei and electrons discovered by Reed Richard