1.

Tube # 3: (a) It started out as purple, and turned into green.
(b) No precipitate was noticably formed.

Tube #4: (a) This tube was purple, and turned into white.
(b) A brown precipitate was formed after adding 1 mL of NaHSO3.

Tube #5: (a) This tube was also purple, but then changed to very light translucent purple.
(b) No prepcipitate was noticably formed.

2. (a) Green is MnO4-, manganate, which is manganese +6.
(b)Brown precipitate is MnO2, manganese dioxide, which is manganese +4, and water-white solutions are Mn2+, manganous.
(c) Purple is MnO4-, permanganate, which is manganese +7.
The HSO3- has sulfur in oxidation state of +4, therefore, it goes to +6 in all of the test tubes.

3. Write the net ionic equations for the redox reactions in each test tube???

I have no idea how to do net ionic equations. All I have are the balanced chemical equations:
(a)2 MnO4- + 2 OH- + SO32- --> 2 MnO42- + SO42- + H2O
(b) 2 MnO4- + 3 HSO3- --> 3 SO42- + H2O + 2 MnO2 + H+
(c) 2 MnO4- + 5 HSO3- + H+ --> 5 SO42- + 3 H2O + 2 Mn2+

Please Help!

a,b, and c are net ionic redox equations. You can't make them any more net ionic than they are.

THANK YOU!!

To write the net ionic equations for the redox reactions in each test tube, you need to identify the species that are undergoing oxidation and reduction.

1. Tube #3:
The color change from purple to green indicates a reduction reaction. The MnO4- ions are being reduced to MnO42- ions. The balanced net ionic equation is:
MnO4- + 4 H2O + 3 e- → MnO42- + 8 OH-

2. Tube #4:
The color change from purple to white, along with the formation of a brown precipitate, suggests a reduction reaction. The MnO4- ions are being reduced to MnO2 solid. The balanced net ionic equation is:
MnO4- + 3 HSO3- + 2 H+ → MnO2 + 3 SO42- + 2 H2O

3. Tube #5:
The color change from purple to light translucent purple indicates a reduction reaction. The MnO4- ions are being reduced to Mn2+ ions. The balanced net ionic equation is:
MnO4- + 8 HSO3- + 5 H+ → 5 SO42- + 3 H2O + 2 Mn2+

These net ionic equations represent the redox reactions happening in each test tube.

To write the net ionic equations for the redox reactions in each test tube, we need to first balance the chemical equations. It seems like you already have the balanced chemical equations, so let's start from there.

(a) 2 MnO4- + 2 OH- + SO32- --> 2 MnO42- + SO42- + H2O

(b) 2 MnO4- + 3 HSO3- --> 3 SO42- + H2O + 2 MnO2 + H+

(c) 2 MnO4- + 5 HSO3- + H+ --> 5 SO42- + 3 H2O + 2 Mn2+

To write the net ionic equations, we need to separate the reactants and products into their respective ions and exclude the spectator ions.

(a) Net Ionic Equation:
2 MnO4- + SO32- + H2O → 2 MnO42- + SO42-

In this net ionic equation, the spectator ions are OH- and H+. They are not involved in the redox reaction, so we remove them.

(b) Net Ionic Equation:
2 MnO4- + 3 HSO3- → 3 SO42- + 2 MnO2 + H2O

In this net ionic equation, the spectator ion is H+. We remove it.

(c) Net Ionic Equation:
2 MnO4- + 5 HSO3- → 5 SO42- + 2 Mn2+ + 3 H2O

In this net ionic equation, there are no spectator ions. All the ions are involved in the redox reaction.

Note: The net ionic equation represents the essential chemical reaction without the spectator ions. It shows the specific ions that are involved in the redox process.

I hope this helps! Let me know if you have any further questions.