Answer:
CO32−
Explanation:
We have to consider the valencies of the polyatomic ions involved. Recall that it is only a polyatomic ion with a valency of -2 that can form a compound which requires two sodium ions.
When we look closely at the options, we will realize that among all the options, only CO32− has a valency of -2, hence it must be the required answer. In order to be double sure, we put down the ionic reaction equation as follows;
2Na^+(aq) + CO3^2-(aq) ---------> Na2CO3(aq)
Acids react with calcium carbonate and more specifically carbonate to form carbon dioxide. An acid will give protons to the carbonate anion to produce carbonic acid which then decomposes into carbon dioxide and water. I hope this helps. Let me know if anything is unclear.
Answer:
K(48.5°C) = 1.017 E-8 s-1
Explanation:
- CH3Cl + H2O → CH3OH + HCl
at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1
at T2 = 48.5°C (321.5 K) ⇒ K2 = ?
Arrhenius eq:
- K(T) = A e∧(-Ea/RT)
- Ln K = Ln(A) - [(Ea/R)(1/T)]
∴ A: frecuency factor
∴ R = 8.314 E-3 KJ/K.mol
⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)
⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)
(1)/(2):
⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)
⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)
⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)
⇒ Ln (K1/K2) = - 3.422
⇒ K1/K2 = e∧(-3.422)
⇒ (3.32 E-10 s-1)/K2 = 0.0326
⇒ K2 = (3.32 E-10 s-1)/0.0326
⇒ K2 = 1.017 E-8 s-1
<u>Answer:</u> The standard potential of the cell is 0.77 V
<u>Explanation:</u>
We know that:
The substance having highest positive 
reduction potential will always get reduced and will undergo reduction reaction.
The half reaction follows:
<u>Oxidation half reaction:</u> 
<u>Reduction half reaction:</u> 
( × 2)
To calculate the 
of the reaction, we use the equation:
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
Putting values in above equation follows:
Hence, the standard potential of the cell is 0.77 V
