Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
Regards.
1. Determine if the ionic substances can break apart into ions.
- e.g. CaCO3 isn't very soluble, do it can't dissolve and dissociate. If it can't pop apart, no ions.
2. Swap the partners for all the other ions that you can get from step 1. You can skip pairings with the same charge - a + can't get close to another + to react.
3. Use solubility, acid/base, and redox rules to see if anything will happen with the ions in solution.<span />
Hard question thx for the points
The answer is B, the number of protons
It's is the unique thing about every element!
