Hey i dont have an answer but i need the points for finals today. Thank you
Answer:
See explanation
Explanation:
All the substances in the mixture; sodium chloride ,lead II oxide and iron filings are solids.
The first step is to use a magnet to separate the iron fillings which is a magnetizable material.
The remaining part of the mixture is now dissolved in water. Lead II oxide is insoluble in water while sodium chloride is very soluble in water hence a filtrate and a residue are obtained.
The residue is lead II oxide which can be dried to recover the solid. The filtrate contains sodium chloride which is recovered by evaporating the solution to dryness.
• Take a look at the steps below to see how to balance this equation. Let's start by writing the unbalanced equation given the information.
Unbalanced Equation : C₃H₈ (g) + O₂ (g) → CO₂ (g) + H₂O (g)
,
Start by Balancing the Carbons : C₃H₈ (g) + O₂ (g) → 3CO₂ (g) + H₂O (g)
Now let's balance the Hydrogen : C₃H₈ (g) + O₂ (g) → 3CO₂ (g) + 4H₂O (g)
Balancing the Oxygen : C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)
Balanced Equation : C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)
• Let's apply dimensional analysis here,
0.7 L propane × (5 liters Oxygen / 1 liter Propane) = 3.5 Liters of Oxygen
• Similarly we can identify the liters of carbon dioxide produced in the reaction,
0.7 L propane × (3 liters Carbon Dioxide / 1 liter Propane) = 2.1 Liters of Carbon Dioxide
• 0.7 L propane × (4 liter water vapor / 1 liter propane ) = 2.8 Liters of Water Vapor
Explanation:
The given data is as follows.
= 286 kJ = 
= 286000 J
, 
Hence, formula to calculate entropy change of the reaction is as follows.
= ![[(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}]](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20S_%7BO_%7B2%7D%7D%29%20-%20%281%20%5Ctimes%20S_%7BH_%7B2%7D%7D%29%5D%20-%20%5B1%20%5Ctimes%20S_%7BH_%7B2%7DO%7D%5D)
= ![[(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)]](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20205%29%20%2B%20%281%20%5Ctimes%20131%29%5D%20-%20%5B%281%20%5Ctimes%2070%29%5D)
= 163.5 J/K
Therefore, formula to calculate electric work energy required is as follows.
= 
= 237.277 kJ
Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.
