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.
Answer:
Saturated solution
We should raise the temperature to increase the amount of glucose in the solution without adding more glucose.
Explanation:
Step 1: Calculate the mass of water
The density of water at 30°C is 0.996 g/mL. We use this data to calculate the mass corresponding to 400 mL.

Step 2: Calculate the mass of glucose per 100 g of water
550 g of glucose were added to 398 g of water. Let's calculate the mass of glucose per 100 g of water.

Step 3: Classify the solution
The solubility represents the maximum amount of solute that can be dissolved per 100 g of water. Since the solubility of glucose is 125 g Glucose/100 g of water and we attempt to dissolve 138 g of Glucose/100 g of water, some of the Glucose will not be dissolved. The solution will have the maximum amount of solute possible so it would be saturated. We could increase the amount of glucose in the solution by raising the temperature to increase the solubility of glucose in water.
Answer:
76.5g KCl/74.55 grams per mole Kcl = x
molality= x/.085 kg H2O
Explanation:
well remember molality is moles of solute/kilograms of solvent. So it's the moles of KCl over 85 g of h20 converted into kg. if this makes sense.
During normal respiration, about 500ml of air enters and leaves the lungs with each respiratory cycle. This is called the<u> tidal volume</u>.
When a person is relaxed, the normal amount of air such a person breathes in and out is called the tidal volume. It is usually measured in millimeters. For the average adult male, it is 500ml, and the tidal volume of average adult female 400ml.
In order to regulate oxygen intake and expulsion of carbon dioxide, the lungs act as buffers in order to absorb the maximum amount of oxygen possible for respiration and other metabolic functions in the body.
When the tidal volume is above or below the 500ml mark, it could signal the presence of underlying pathological conditions like bronchitis, emphysema and asthma.
To find out more about tidal volume, visit:
brainly.com/question/17439101
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