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
six commonly recognised metalloids are boron, silicon, germanium, arsenic, antimony, and tellurium. Five elements are less frequently so classified: carbon, aluminium, selenium, polonium, and astatine.
J.j Thomson electrons demo true indivisible sphere n Ernest central nucleas
Answer:
66 g of CO₂
Solution:
The Balance Chemical Reaction is as follow,
C₂H₂ + 5/2 O₂ → 2 CO₂ + H₂O
Or,
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O ------- (1)
Step 1: Find out the limiting reagent as;
According to Equation 1,
56.1 g (2 mole) C₂H₂ reacts with = 160 g (5 moles) of O₂
So,
125 g of C₂H₂ will react with = X g of O₂
Solving for X,
X = (125 g × 160 g) ÷ 56.1 g
X = 356.5 g of O₂
It means for total combustion of Ethylene we require 356.5 g of O₂, but we are only provided with 60.0 g of O₂. Therefore, O₂ is the limiting reagent and will control the yield.
Step 2: Calculate Amount of CO₂ produced as;
According to Equation 1,
160 g (5 mole) O₂ produces = 176 g (4 moles) of CO₂
So,
60.0 g of O₂ will produce = X g of CO₂
Solving for X,
X = (60.0 g × 176 g) ÷ 160 g
X = 66 g of CO₂
