Where is each conversion factor found?
1 answer:
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Answer:
Carboxylic acid
A carboxylic acid is an organic acid that contains a carboxyl group attached to an R-group. The general formula of a carboxylic acid is R−COOH or R−CO₂H, with R referring to the alkyl, alkenyl, aryl, or other group. Carboxylic acids occur widely. Important examples include the amino acids and fatty acids.
There are:
3.41 moles of C
4.54 moles of H
3.40 moles of O.
Why?
To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.
Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.
We have that:
To know the percent of each element, we need to to the following:
So, we know that for the 100 grams of the compound, there are:
40.92 grams of C
4.58 grams of H
54.49 grams of O
We know the molecular masses of each element:
Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:
Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.
Have a nice day!
Answer:
Explanation:
The cell reaction properly written is shown below:
Cu|Cu²⁺ || Ag⁺
| Ag
From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.
Oxidation half:
Cu ⇄ Cu²⁺
+ 2e⁻
At the anode, oxidation occurs.
Reduction half:
Ag⁺ + 2e⁻ ⇄ Ag
At the cathode, reduction occurs.
To derive the overall reaction, we must balance the atoms and charges:
Cu ⇄ Cu²⁺
+ 2e⁻
Ag⁺ + e⁻ ⇄ Ag
we multiply the second reaction by 2 to balance up:
2Ag⁺ + 2e⁻ ⇄ 2Ag
The net reaction equation:
Cu + 2Ag⁺
+ 2e⁻⇄ Cu²⁺
+ 2e⁻ + 2Ag
We then cancel out the electrons from both sides since they appear on both the reactant and product side:
Cu + 2Ag⁺
⇄ Cu²⁺
+ 2Ag