Ionization energy (IE) is the amount of energy required to remove an electron.
If you observe the IEs sequentially, there is a large gap between the 2nd and 3rd. This suggests it is difficult to remove more than 2 two electrons. Elements that lose two electrons to become more stable are found in the Group 2A (2 representing the number of electrons in the outermost valence shell).
The chloroplasts i believe is the answer
Answer:
2H+(aq) + 2OH-(aq) → 2H2O(l)
Explanation:
Step 1: The balanced equation
2HCl(aq)+Ca(OH)2(aq) → 2H2O(l)+CaCl2(aq)
This equation is balanced, we do not have the change any coefficients.
Step 2: The netionic equation
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will.
2H+(aq) + 2Cl-(aq) + Ca^2+(aq) + 2OH-(aq) → 2H2O(l) + Ca^2+(aq) + 2Cl-(aq)
After canceling those spectator ions in both side, look like this:
2H+(aq) + 2OH-(aq) → 2H2O(l)
Answer:
Other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide. However iodides and bromides have different colours hence they will not give a false positive test for KCl. Other chlorides present may lead to a false positive test for KCl.
Explanation:
In the qualitative determination of halogen ions, silver nitrate solution is used. Various halide ions give various colours of precipitate with silver nitrate. Chlorides yield a white precipitate, bromides yield a cream precipitate while iodides yield a yellow precipitate. All these ions or some of them may be present in the system.
However, if other chlorides are present, they will also yield a white precipitate just as KCl leading to a false positive test for KCl. Since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. We can exclude other halides from the tendency to lead us to a false positive test for KCl but not other chlorides.
