For the first one it’s 69 just count the little lines from the side.
Answer:
202 L
Explanation:
Step 1: Write the balanced equation
C₆H₁₂O₆ + 6 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(l)
Step 2: Calculate the moles corresponding to 270 g of C₆H₁₂O₆
The molar mass of C₆H₁₂O₆ is 180.16 g/mol.
270 g × 1 mol/180.16 g = 1.50 mol
Step 3: Calculate the moles of CO₂ generated from 1.50 moles of glucose
The molar ratio of C₆H₁₂O₆ to CO₂ is 1:6. The moles of CO₂ formed are 6/1 × 1.50 mol = 9.00 mol
Step 4: Calculate the volume of 9.00 moles of CO₂ at STP
The volume of 1 mole of an ideal gas at STP is 22.4 L.
9.00 mol × 22.4 L/mol = 202 L
I would think that b would be the right answer
Boiling point elevation is given as:
ΔTb=iKbm
Where,
ΔTb=elevation in the boiling point
that is given by expression:
ΔTb=Tb (solution) - Tb (pure solvent)
Here Tb (pure solvent)=118.1 °C
i for CaCO3= 2
Kb=2.93 °C/m
m=Molality of CaCO₃:
Molality of CaCO₃=Number of moles of CaCO₃/ Mass of solvent (Kg)
=(Given Mass of CaCO3/Molar mass of CaCO₃)/ Mass of solvent (Kg)
=(100.0÷100 g/mol)/0.4
= 2.5 m
So now putting value of m, i and Kb in the boiling point elevation equation we get:
ΔTb=iKbm
=2×2.93×2.5
=14.65 °C
boiling point of a solution can be calculated:
ΔTb=Tb (solution) - Tb (pure solvent)
14.65=Tb (solution)-118.1
Tb (solution)=118.1+14.65
=132.75
