Answer:
Incomplete precipitation of barium sulfate
Explanation:
The student has precipitated and digested the barium sulfate on his/her side. But on the addition of 
in the solution, the solution become cloudy. This happened because incomplete precipitation of barium sulfate by the student. When is added, there are still sulfate ions present in the solution with combines with and forms 
and the formation of this precipitate makes the solution cloudy.
Answer:
The balanced chemical equation: 2 Al + 3Cl2→ 2 AlCl3
Mole-mole relationship: 2 moles Al + 3 moles Cl2→ 2 moles AlCl3
Given: 0.600 moleCl2; 0.500 mole Al
Required: Excess reactant___; Number of moles of AlCl3 produced__
Solution: Use dimensional analysis using the mole-mole rel
0.600 mole Cl2 * 2 moles Al/3 moles Cl2 = 0.4 mole Al
0.5 mole Al* 3 moles Cl2/2 moles Al = 0.75 mole Cl2
Based on the given:
0.6mole Cl2 + 0.4 mole Al ( this is possible based on the given)
0.5mole Al + 0.75 mole Cl2 (this is not possible because the given is only 0.600 mole of Cl 2)
Answer: Excess reactant is Al; Limiting reactant is Cl2
The amount of AlCl3 produced = 0.6 mole Cl2 + 0.4 mole Al = 1.0 mole AlCl3
Answer:
Well this has a good chance of being wrong but i wanna say,
When you change a physical property of something it doesn't affect the chemicals that make it up. Like Ice, you can freeze water to make ice. You change a physical property (state of matter) but it's chemical properties don't change because in the end it's still water.
However if you remove a chemical property from something you are changing what made the new substance with will also change the substance along with it.
That's just what I think though
Explanation:
Answer:
= 160 mL
Explanation:
The equation for the reaction is;
Ca(OH)2 + 2HNO3 -----> Ca(NO3)2 + 2H2O
Moles of Ca(OH)2 = 0.080 L x 5.00^-3 M
= 0.0004 moles
From the reaction; 2 moles of HNO3 reacts with 1 mole of Ca(OH)2
Thus; moles of HNO3 = 0.0004 Moles × 2
= 0.0008 moles
Volume = moles/ molarity
Therefore; volume of HNO3= 0.0008 Moles/ 5 × 10^-3 M
<u>= 0.16 L or 160 mL</u>
