<span>
Mn²⁺ + 4H2O -----> MnO4⁻ + 8H⁺ +5e⁻ /*2
<span>NaBiO3 +6H⁺ +2e⁻ -----> Bi³⁺ + Na⁺ + 3H2O /*5
</span>2Mn²⁺ + 5 NaBiO3+8H2O+30H⁺ ---> 2MnO4⁻ +5Bi³⁺ + 5Na⁺ +16H⁺ +15H2O
</span>2Mn²⁺ + 5 NaBiO3+14H⁺ ---> 2MnO4⁻ +5Bi³⁺ + 5Na⁺ +7H2O
There are 7 water molecules in this reaction.
Answer:
-The other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide.
-It is reasonable to exclude iodides and bromides but it is not reasonable to exclude other chlorides
Explanation:
In the qualitative determination of halogen ions, silver nitrate solution(AgNO3) is usually used. Now, various halide ions will give various colours of precipitate when mixed with with silver nitrate. For example, chlorides(Cl-) normally yield a white precipitate, bromides(Br-) normally yield a cream precipitate while iodides (I-) normally yield a yellow precipitate. Thus, all these ions or some of them may be present in the system.
With that being said, if other chlorides are present, they will also yield a white precipitate just like KCl leading to a false positive test for KCl. However, since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. Thus, we can exclude other halides from the tendency to give us a false positive test for KCl but not other chlorides.
Answer:
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A starfish has five equal arms, so a starfish would have 5 lines of symmetry. Hope this helps!
<span> When an </span>acid and a base<span> are placed together, they </span>react<span> to neutralize the </span>acid<span> and </span>base<span> properties, producing a salt. The H(+) cation of the </span>acid<span>combines with the OH(-) anion of the </span>base<span> to form water.</span>
