Answer:
Noble gases.
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Answer:
pH = 9.48
Explanation:
We have first to realize that NH₃ is a weak base:
NH₃ + H₂O ⇔ NH₄⁺ + OH⁻ Kb = 1.8 x 10⁻⁵
and we are adding this weak base to a solution of NH₄NO₃ which being a salt dissociates 100 % in water.
Effectively what we have here is a buffer of a weak base and its conjugate acid. Therefore, we need the Henderson-Hasselbach formula for weak bases given by:
pOH = pKb + log ( [ conjugate acid ] / [ weak base ]
mol NH₃ = 0.139 L x 0.39 M = 0.054 mol
mol NH₄⁺ = 0.169 L x 0.19 M = 0.032 mol
Now we have all the information required to calculate the pOH ( Note that we dont have to calculate the concentrations since in the formula they are a ratio and the volume will cancel out)
pOH = -log(1.8 x 10⁻⁵) + log ( 0.032/0.054) = 4.52
pOH + pH = 14 ⇒ pH = 14 - 4.52 = 9.48
The solution is basic which agrees with NH₃ being a weak base.
Answer:
Ammonium chloride
Explanation:
The powder is:- Ammonium chloride
When mixed with silver nitrate, white prescipitate of silver chloride is formed as:-
When mixed with sodium hydroxide, ammonia gas is formed which has noxious order.
Ammonia gas on reaction with nickel (II) hydroxide forms deep blue colored complex as shown below:-
![Ni(OH)_2(s) + 6NH_3(aq)\rightarrow [Ni(NH_3)_6]^{2+}(aq) + 2OH^{-}(aq)](https://tex.z-dn.net/?f=Ni%28OH%29_2%28s%29%20%2B%206NH_3%28aq%29%5Crightarrow%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%28aq%29%20%2B%202OH%5E%7B-%7D%28aq%29)
Answer:
179.0 g of iridium (1 mol / 192.217 g) ( 6.022 x 10^23 atoms / 1 mol ) = 5.61 x 10^23 atoms of iridium
Explanation:
True the answer is true :) have a nice day
