Answer:
1.18 × 10⁷ c
Iron is the anode and zinc is the cathode.
Explanation:
Let's consider the reduction of Zn²⁺.
Zn²⁺(aq) + 2 e⁻ ⇒ Zn(s)
<em>How many coulombs of charge are needed to produce 61.2 mol of solid zinc?</em>
<em />
We can establish the following relations:
- When 2 moles of electrons circulate, 1 mol of Zn is produced.
- 1 mole of electrons have a charge of 96468 c (Faraday's constant).
Then, for 61.2 mol of Zn:
<em>Identify the anode and cathode when plating an iron nail with zinc.</em>
The anode is where the oxidation takes place and the cathode is where the reduction takes place.
Anode (oxidation): Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode (reduction): Zn²⁺(aq) + 2 e⁻ ⇒ Zn(s)
Answer
is: activation energy of this reaction is 212,01975 kJ/mol.<span>
Arrhenius equation: ln(k</span>₁/k₂) =
Ea/R (1/T₂ - 1/T₁).<span>
k</span>₁
= 0,000643 1/s.<span>
k</span>₂
= 0,00828 1/s.
T₁ = 622 K.
T₂ = 666 K.
R = 8,3145 J/Kmol.
<span>
1/T</span>₁ =
1/622 K = 0,0016 1/K.<span>
1/T</span>₂ =
1/666 K = 0,0015 1/K.<span>
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol ·
(-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol </span>· (-0,0001 1/K).<span>
Ea = 212019,75 J/mol = 212,01975 kJ/mol.</span>
The concentrations : 0.15 M
pH=11.21
<h3>Further explanation</h3>
The ionization of ammonia in water :
NH₃+H₂O⇒NH₄OH
NH₃+H₂O⇒NH₄⁺ + OH⁻
The concentrations of all species present in the solution = 0.15 M
Kb=1.8 x 10⁻⁵
M=0.15
![\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}](https://tex.z-dn.net/?f=%5Ctt%20%5BOH%5E-%5D%3D%5Csqrt%7BKb.M%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-5%7D%5Ctimes%200.15%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B2.7%5Ctimes%2010%5E%7B-6%7D%7D%3D1.64%5Ctimes%2010%5E%7B-3%7D)
![\tt pOH=-log[OH^-]\\\\pOH=3-log~1.64=2.79\\\\pH=14-2.79=11.21](https://tex.z-dn.net/?f=%5Ctt%20pOH%3D-log%5BOH%5E-%5D%5C%5C%5C%5CpOH%3D3-log~1.64%3D2.79%5C%5C%5C%5CpH%3D14-2.79%3D11.21)
Answer: 
Explanation:
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
Here metal X is having an oxidation state of +4 called as 
cation and 
is an anion with oxidation state of -3. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral
is the formula of the resulting compound.
