Answer:
See explanation.
Explanation:
A. 2s is incorrect, the right arrow should be pointing downwards.
B. In 2p, the last box has 3 arrows. This is invalid. Each box can only store 2 arrows(electrons). The last arrow should be moved to a 3p orbital (new box).
C. In this one, the arrows switched directions (left is down and right is up). You could fix each box or just the last one (in 3p) to point downwards.
There would be a direct result as an increase in the solute temperature will result in increase in its solubility. A greater amount of solute molecules will possess more kinetic energy and will be distributed and in container.
The Balanced Chemical Equation is as follow;
4 KO₂ + 2 CO₂ → 2 K₂CO₃ + 3 O₂
First find out the Limiting Reagent,
According to equation,
284 g (4 moles) KO₂ reacted with = 44.8 L (2 moles) of CO₂
So,
27.9 g of KO₂ will react with = X L of CO₂
Solving for X,
X = (44.8 L × 27.9 g) ÷ 284 g
X = 4.40 L of CO₂
Hence, to consume 27.9 g of KO₂ only 4.40 L CO₂ is required, while, we are provided with 29 L of CO₂, it means CO₂ is in excess and KO₂ is is limited amount, Therefore, KO₂ will control the yield of K₂CO₃. So,
According to eq.
284 g (4 moles) KO₂ formed = 138.2 g of K₂CO₃
So,
27.9 g of KO₂ will form = X g of K₂CO₃
Solving for X,
X = (138.2 g × 27.9 g) ÷ 284 g
X = 13.57 g of K₂CO₃
So, 13.57 g of K₂CO₃ formed is the theoretical yield.
%age Yield = 13.57 / 21.8 × 100
%age Yield = 62.24 %
