We can solve this without a concrete formula through dimensional analysis. This works by manipulating the units such that you end up with the unit of the final answer. Manipulate them by cancelling units that appear both in the numerator and denominator side. As a result, we must be left with the units of g. The current in A or amperes is equivalent to amount of Coulombs per second. Since this involves Coulombs, we will use the Faraday's constant which is 96,500 C/mol electron. The reaction is:
Cr³⁺(aq) + 3e⁻ --> Cr(s)
This means that for every 3 moles of electron transferred, 1 mole of Chromium metal is plated. The molar mass of Cr: 52 g/mol. The solution is as follows:
Mass of Chromium metal = (8 C/s)(60 s/1 min)(160 min)(1 mol e⁻/96,500 C)(1 mol Cr/3 mol e)(52 g/mol)
<em>Mass of Chromium metal = 13.79 g</em>
Well, first we must remember that

This is because


So then

Answer: The concentration of C29H60 in nM per liter is 83,33 nM/liter
Explanation: Let's start from the ppb definition: ppb means parts per billion. In terms of concentracion measuring this means micrograms of solute per liter of solution.
The algebraic expression would be:
<em>ppb [=] micrograms of compound/liter of solution</em>
We can assume that the solvent is water. The solute is dissolved in water and both create the C29H60 solution.
For the exercise we have 34 ppb of C29H60, that means 34 micrograms of C29H60 in one liter of solution. So, since now, we have to convert the units from the initial data to the required answer.
The respective procedure is in a attached file.
Hi, you have not provided structure of the aldehyde and alkoxide ion.
Therefore i'll show a mechanism corresponding to the proton transfer by considering a simple example.
Explanation: For an example, let's consider that proton transfer is taking place between a simple aldehyde e.g. acetaldehyde and a simple alkoxide base e.g. methoxide.
The hydrogen atom attached to the carbon atom adjacent to aldehyde group are most acidic. Hence they are removed by alkoxide preferably.
After removal of proton from aldehyde, a carbanion is generated. As it is a conjugated carbanion therefore the negative charge on carbon atom can conjugate through the carbonyl group to form an enolate which is another canonical form of the carbanion.
All the structures are shown below.