Answer:
The reaction rate becomes quadruple.
Explanation:
According to the law of mass action:-
The rate of the reaction is directly proportional to the active concentration of the reactant which each are raised to the experimentally determined coefficients which are known as orders. The rate is determined by the slowest step in the reaction mechanics.
Order of in the mass action law is the coefficient which is raised to the active concentration of the reactants. It is experimentally determined and can be zero, positive negative or fractional.
The order of the whole reaction is the sum of the order of each reactant which is raised to its power in the rate law.
Thus,
Given that:- The rate law is:-
![r=k[A_2][B_2]](https://tex.z-dn.net/?f=r%3Dk%5BA_2%5D%5BB_2%5D)
Now, ![[A'_2]=2[A_2]](https://tex.z-dn.net/?f=%5BA%27_2%5D%3D2%5BA_2%5D)
and ![[B'_2]=2[B_2]](https://tex.z-dn.net/?f=%5BB%27_2%5D%3D2%5BB_2%5D)
So, ![r'=k[A'_2][B'_2]=k\times 2[A_2]\times 2[B_2]=4\times k[A_2][B_2]=4r](https://tex.z-dn.net/?f=r%27%3Dk%5BA%27_2%5D%5BB%27_2%5D%3Dk%5Ctimes%202%5BA_2%5D%5Ctimes%202%5BB_2%5D%3D4%5Ctimes%20k%5BA_2%5D%5BB_2%5D%3D4r)
<u>The reaction rate becomes quadruple.</u>
Reduction half reaction: Cu²⁺(aq) + 2e⁻ → Cu⁰(s).
Oxidation half reaction: NO₂⁻(aq) + H₂O(l) → NO₃⁻(aq) + 2H⁺(aq) + 2e⁻.
Balanced chemical reaction:
Cu²⁺(aq) + NO₂⁻(aq) + H₂O(l) → Cu(s) + NO₃⁻(aq) + 2H⁺(aq).
Copper is reduced from oxidation number +2 (Cu²⁺) to oxidation number 0 (Cu) and nitrogen is oxidized from oxidation number +3 (in NO₂⁻) to oxidation number +5 (in NO₃⁻).
soluble substances dissolves in solvents
Answer:
I hate this i already know all of you are using this site
Explanation:
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this is a sign
Explanation:
1) Amphoteric substances can react both like an acid or a base depending if they are in presence of a base or an acid.
Zinc hydroxide (Zn(OH)2):
In acid: 
In base: ![Zn(OH)_2 + 2 NaOH \longrightarrow Na_2[Zn(OH)_4]](https://tex.z-dn.net/?f=Zn%28OH%29_2%20%2B%202%20NaOH%20%5Clongrightarrow%20Na_2%5BZn%28OH%29_4%5D)
Aluminium hydroxide (Al(OH)3)
In acid: 
In base: ![Al(OH)_3 + NaOH \longrightarrow Na[Al(OH)_4]](https://tex.z-dn.net/?f=Al%28OH%29_3%20%2B%20NaOH%20%5Clongrightarrow%20Na%5BAl%28OH%29_4%5D)
2) Reactions
a) 
b) 
