Some troubling problems...

[Predentknight]

---

Members do not see this ad.

Some problems I am having trouble with:

1. Which of the following descriptions relate to nuclear reactions involving one beta emission. I.Transformation of a nuclear proton into a nuclear neutron II.Transformation of a nuclear neutron into a nuclear protonIII.Emission of one electron

Whats the ans. and why?

2.Balance the following redox equation and determine the sum of coefficients for both the reactants and the products.
Al(s) + NO3-(aq) + H+(aq) à N2O4(g) + H2O(l) + Al3+(aq)

Lost


I will post answers once I get some responses

 

[doc3232]

For the second one

I got 2 NO3 4 H 1 Al
Right side: 1 N2O4 2 H2O and 1 Al
The electrons should be set correctly.

 

[osimsDDS]

for #1: is the answer II and III?

so basically when you undergo B emission you are adding 1 prton and 0 neutrons. So you would have lets say C-12 with 6 protons...you are adding a proton so you would have 7 protons. Now take the difference to see how many neutrons you have: 12-7= 5, and before you have 21-6= 6. So you went from 6 neutrons to 5 neutrons so 1 of those neutrons had to be changed into a proton...now for answer III, when it is B positron emission it is called B + decay and when its B electron emission it is called B- Decay...you just have to know that when you have beta emission it emitts an electron and when its B+ emission or positron emission it emitts a positron (e+).

Hope thats right haha

 

[happyasaclam88]

when i did the redox, i got a lot messier numbers in order to get the electrons to even out. i got:
2 Al + 12 H+ + 6 NO3- for reactants
3 N2O4 + 2 Al3+ + 6H2O for products.

what is the answer? thanks!

 

[osimsDDS]

I got:

two half reaction are:
Al----> Al3+ + 3e-
2e- + 2NO3- + 4H+ ----> N2O4 + 2H20

then you multiply the bottom half reaction by 3 and the top by 2 to cancel out the electrons then add them up and you should get:

Reactants: 2Al, 6 NO3-, 12H+
Products: 3 N2O4, 6 H20, and 2 Al3+

 

[Predentknight]

for #1: is the answer II and III?

so basically when you undergo B emission you are adding 1 prton and 0 neutrons. So you would have lets say C-12 with 6 protons...you are adding a proton so you would have 7 protons. Now take the difference to see how many neutrons you have: 12-7= 5, and before you have 21-6= 6. So you went from 6 neutrons to 5 neutrons so 1 of those neutrons had to be changed into a proton...now for answer III, when it is B positron emission it is called B + decay and when its B electron emission it is called B- Decay...you just have to know that when you have beta emission it emitts an electron and when its B+ emission or positron emission it emitts a positron (e+).

Hope thats right haha

This is the correct answer to #1, that really helps,thanks for the explanation


And for #2 the ans. is 31. Happyasaclam88 has the right answer do you mind explaining how you come to that.