Zinc would be considered the strongest reducing agent.
<h3>Reducing agent</h3>
A reducing agent is a chemical species that "donates" one electron to another chemical species in chemistry (called the oxidizing agent, oxidant, oxidizer, or electron acceptor). Earth metals, formic acid, oxalic acid, and sulfite compounds are a few examples of common reducing agents.
Reducers have excess electrons (i.e., they are already reduced) in their pre-reaction states, whereas oxidizers do not. Usually, a reducing agent is in one of the lowest oxidation states it can be in. The oxidation state of the oxidizer drops while the oxidizer's oxidation state, which measures the amount of electron loss, increases. The agent in a redox process whose oxidation state rises, which "loses/donates electrons," which "oxidizes," and which "reduces" is known as the reducer or reducing agent.
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Answer:
The net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide is 523.2 kJ.
Explanation:
Step 1:
...[1]
Step 2 :
..[2]
Adding 6 × [1] and [2]:


we get :




Energy released on formation of 5 moles of acrylic acid = 2,626 kJ
Energy released on formation of 1 mole of acrylic acid:

Answer:
See explanation and picture below
Explanation:
First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.
In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.
In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.
Answer:
Acid(BSA) = CH₃COOH
Base (BSB) = H₂O
Conjugate base (CB) = CH₃COO⁻
Conjugate acid (CA) = H₃O⁺
Explanation:
Equation of reaction;
CH₃COOH + H₂O → CH₃COO⁻ + H₃O⁺
Hello,
From my understanding of the question, we are required to identify the
1) Acid
2) Base
3) conjugate acid
4) conjugate base in the reaction
Acid (BSA) = CH₃COOH
Base (BSB) = H₂O
CA = conjugate acid = H₃O⁺
CB = conjugate base = CH₃COO⁻