Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
<u>Answer:</u>
<em>Atomic number 75 is dedicated to an element named rhenium and has been given Re as its chemical name.</em>
<u>Explanation:</u>
With a really low concentration it is one of the rarest metals that is found in Earth's crust.
Like all other elements rhenium also has certain isotopes among with 185 and 187 are the most stable ones. Hence these two are the ones that are naturally available abundance is 34% and 63% respectively.
Answer: I just took the test. The answer is D! (A single replacement reaction takes place because sodium is more reactive than hydrogen.)
Answer:
1= 2H₂ + O₂ → 2H₂O
2=CaCo₃ + heat → CaO +CO₂
3=CH₄ + 2O₂ → CO₂ +2H₂O
4=HCl + NaOH → NaCl + H₂O
Explanation:
1 = Simple composition
The formation of water molecule is simple composition reaction. In this reaction two hydrogen atoms react with one oxygen atom and form one water molecules.
2H₂ + O₂ → 2H₂O
The amount of energy released is -285.83 KJ/mol. It is exothermic reaction.
2 = Simple decomposition reaction:
The break down of sodium hydrogen carbonate into sodium carbonate, carbondioxide and water is decomposition reaction. The decomposition reactions re mostly endothermic, because compound required energy to break.
2NaHCO₃ + heat → Na₂CO₃ + H₂O + CO₂
It is endothermic reaction.
Another example is:
CaCo₃ + heat → CaO +CO₂
3 = Combustion reaction
Consider the combustion of methane:
CH₄ + 2O₂ → CO₂ +2H₂O
The burning of methane is exothermic. The combustion reactions are exothermic because when fuel are burns they gives energy.
4 = Neutralization reaction
The neutralization reactions are those in which acid and base react to form the salt and the water. Some neutralization reactions are exothermic because they release heat. e.g
Consider the neutralization reaction of HCl and NaOH.
HCl + NaOH → NaCl + H₂O
The dissociation of formic acid is:
The acid dissociation constant of formic acid, 
is:
![k_a = \frac{[HCOO^{-}] [H^{+}]}{HCOOH}](https://tex.z-dn.net/?f=%20k_a%20%3D%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%20%20%5BH%5E%7B%2B%7D%5D%7D%7BHCOOH%7D%20%20%20%20%20)
Rearranging the equation:
![\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%7D%7B%5BHCOOH%5D%7D%20%3D%20%5Cfrac%7Bk_a%7D%7B%5BH_%2B%5D%7D%20)
pH = 2.75
![pH = -log[H^{+}]](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%5BH%5E%7B%2B%7D%5D%20)
![[H^{+}]= 10^{-2.75} = 1.78 \times 10^{-3}](https://tex.z-dn.net/?f=%20%5BH%5E%7B%2B%7D%5D%3D%2010%5E%7B-2.75%7D%20%3D%201.78%20%5Ctimes%2010%5E%7B-3%7D%20)
Substituting the values in the equation:
![\frac{[HCOO^{-}]}{[HCOOH]} = \frac{1.78\times 10^{-4}}{1.78\times 10^{-3}}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%7D%7B%5BHCOOH%5D%7D%20%3D%20%5Cfrac%7B1.78%5Ctimes%2010%5E%7B-4%7D%7D%7B1.78%5Ctimes%2010%5E%7B-3%7D%7D%20%20%20)
Hence, the ratio is 
.
