Which of the following statements is true for most atoms? In general, as the ato

Question

Which of the following statements is true for most atoms? In general, as the atomic radius increases, the first ionization energy increases. In general, as the atomic radius increases, the first ionization energy decreases. In general, as the first ionization energy decreases, the electronegativity increases. No correct answer is given. Draw the best Lewis structure for NO_2^- Balance the following equations which occur in basic solution. MnO_4^3- + l^-3 rightarrow MnO_2 + IO_2^- H_2O_2 + MnO_4 rightarrow O_2 + MnO_2 Balance the following redox reaction is acidic solution and calculate the mass of iodine produced from the reaction of 100.0 mL of an aqueous solution that is 0.125 M in I^- with 100.0 mL of a an aqueous solution that is 0.125 M in NO_2^-. For the following neutralization reaction; hydrochloric acid plus barium hydroxide yields barium chloride plus water. Write the balanced formula equation. Write the total ionic equation. Write the net ionic equation. A 400.g sample of Aluminium sulfate is dissolved in 3000. mL of solution. What is the molarity of the solution? What is the molarity of aluminium ion in the solution? What is the molarity of the sulfate ion in the solution? A write complete electron configurations for Aluminium, nickel, and rubidium.

Explanation / Answer

9)

First we want to split this equation into two relative equations. One will be the oxidation part of the equation, and the other will focus on the reduction.

To identify the oxidation equation you should first write the equation in ionic form to identify which element is being reduced and which is being oxidized. In this case, the Manganese is being reduced, and the Iodine is being oxidized.

To explain this, in the reactants, the Manganese has an oxidation number of +7. It has to have this charge to counteract the 8- from the oxygen (Again, if you need this explained as well, I will include it at the bottom of my answer). The iodine is being oxidized because it goes from having a charge of 1- to a charge of 5+ (It has a 5+ charge due to the 6- charge of the oxygen).

Now that we have done this, the oxidation equation would be:

IIO3

And the reduction equation would be:

MnO4MnO2

In this equation, both the reduction and oxidation equations have oxygen in them. Because of this we must add molecules of water to the side of the equation that needs oxygen to balance.

However, once we add water, since water also consists of hydrogen, we must add H+ ion to whichever side is lacking them. I think its easier if I just show you:

3H2O+ I IO3

Here we added three molecules of H2O because that will give us the three atoms of oxygen that the product side of this equation has. But... we are still missing something. Now the reactant side has six extra hydrogens that the product side does not.

When dealing with this step in the balancing process, it is best to remember:

Now that we've covered the oxygen, we must cover the hydrogen part of this:

3H2O+I IO3+ 6H+

One more thing needs to be done here. Ask yourself, is the net charge on both sides of the equation equal? The answer is no, they are not.

To fix this issue, you must add a negative charge to the equation to balance the charges. You do this by adding electrons. For this equation, the left side already has a net charge of 1-. We cannot change this, or make it positive. Remember:

You can only add a negative charge.

This is what you would do:

3H2O+ I IO3 + 6H+ + 6e

Good now both sides have a charge of 1- and the oxidation equation is balanced.

On to reduction: Here you do the same thing. I am just going to give you the balanced equation here, and if you need reference, it is the exact same process that we did to the oxidation part of the equation:

3e+ 4H+ +MnO4 MnO2+2H2O

Now that we have both equations balanced, we do something similar to the process of elimination in math equations.

(3H2O+IIO3+ 6H+ +6e)

(3e+4H++MnO4MnO2+2H2O)2

We multiply the second equation by two so that:

*The electrons on both equations are equal. *

This means that we multiplied by two because the first equation has six electrons while the second only has three. (3e2=6e)

Now rewrite what we have:

3H2O+ IIO3+ 6H+ + 6e

6e+ 8H+ +2MnO4 2MnO2+4H2O

We can now cancel out similarities so we are left with:

2H++ I +2MnO42MnO2+ IO3+ H2O

Finally, if this were an acidic solution, we would be done and this would be the correct answer. However, we are in a basic solution.

So since we are, we must add OH ions, or hydroxide ions, to replace the H+ions. We add 2 OH ions to both sides of the equation because there are 2H+ions.

When you add them to the reactant side of the equation, you get two molecules of H2O, and on the other side, you get an additional 2OH.

Here:

2H2O+I+2MnO42MnO2+IO3+ H2O+ 2OH

The similar waters cancel out and...

H2O+ I+ 2MnO4 2MnO2+ IO3+ 2OH

We are now done with the equation, and the above equation would be your final answer.

b) Similarly we get,

2 MnO4- + 3 H2O2 ---> 3 O2 + 2 MnO2 + 2 H2O + 2OH-

10) 2I- + NO2- ---> I2 + NO2 + 3e-  

given , no. of moles of I- = 0.1L * 0.125 M = 0.0125

no. of moles of NO2- = 0.1L*0.125M = 0.0125

=> no. of moles of iodine produced = 0.0125/2 = 0.00625

Mass of iodine produced = no. of moles * Molecular weight = 0.00625*253.81 = 1.586 gm

11)  Ba(OH)2(s) + 2HCl(aq) --> 2H2O(aq) + BaCl2(aq)

A neutralization equation is when an acid and base react to form water and a salt... this is ALWAYS true.

To write the net ionic equation, break the molecules in aqueous forms into their ions and if they appear on both sides of the equation, cancel them out. Then rewrite the equation.

Thus,
Ba(OH)2(s) + 2HCl(aq) --> 2H2O(l) + BaCl2(aq)
Total ionic equation:

Ba(OH)2(s) + 2H+ + 2Cl- --> 2H2O(l) + Ba^2+ + 2Cl-

Net ionic equation:

Ba(OH)2(s) 2H+(aq) --> 2H2O(l) + Ba^2+(aq)

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