<span>The density of the solution =1.05 g/ml.
</span><span>The total mass of the resulting solution is = 398.7 g (CaCl2 + water)
</span>
Find moles of CaCl2 and water.
Molar mass of CaCl2 = 110 (approx.)
Moles of CaCl2 = 23.7 / 110 = 0.22
so, moles of Cl- ion = 2 x 0.22 = 0.44 (because each molecule of CaCl2 will give two Cl- ions)
Moles of water = 375 / 18 = 20.83
Now, Mole fraction of CaCl2 = (moles of CaCl2) / (total moles)
total moles = moles of Cl- ions + moles of Ca2+ ions + moles of water
= 0.44 + 0.22 + 20.83
=21.49
So, mole fraction = 0.44 / (21.49) = 0.02
Guess what !!! density is not used. No need
Some of the muscle attached to the skeleton is voluntary and may be used for movement.
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
Answer:
do you know the answer to it
Explanation:
