Answer:
Hi.
The temperature is approximately zero degrees (0°C)
Explanation:
It is important to keep in mind that in the production of ice cream the decrease in the freezing point of the water present in the mixture is called the antifreeze power of the mixture. In ice cream, the freezing point decrease will be caused by each substance that is dissolved in the mixture: lactose, salts, sugars and any other substance. Each of these substances will contribute to the decrease in the freezing point of the mixture. The phase diagram attached in the file shows the sugar solutions in water. When a solution cools (point A), there comes a time when the freezing curve is reached (point B). At that moment ice begins to appear. As shown in the diagram this temperature is approximately zero degrees (0 ° C).
A high concentration of water has <u>fewer</u> dissolved particles than a low water concentration.
Most cell membranes are not as easily permeable to many dissolved compounds as water is. There is a quick and constant flow of water. From one area with less dissolved matter to another with more, water transports NET. Or, if you want, from an area with a lot of water to one with little water. The terms isotonic, hypotonic, and hypertonic refer to the concentration of dissolved material. In a medium, such as the extracellular fluid, every distinct material has a concentration gradient that is unique from the gradients of other substances. Every substance will diffuse in line with that gradient as well.
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The answer is iii) decreasing the pressure of the system. When the pressure is decreased, the equilibrium will shift to the right because it has 12 moles of gas which is greater than the number of moles of gas on the left side, which is 6 moles. Equilibrium shifting to the side that exerts greater pressure is favored to offset the decrease in pressure.
Answer:
Product: ethyl L-valinate
Explanation:
If we want to understand what it is the molecule produced we have to an<u>alyze the reagents</u>. We have valine an <u>amino acid</u>, in this kind of compounds we have an <em>amine group</em> (
) and a <em>carboxylic acid</em> group (
). Additionally, we have an <u>alcohol </u>(
) in the presence of HCl (a <u>strong acid</u>) in the first step, and a base (
).
When we have an acid and an alcohol in a vessel we will have an <u>esterification reaction</u>. In other words, an ester is produced. As the <em>first step,</em> the oxygen in the C=O (in the carboxylic acid group) would be protonated. In the <em>second step</em>, the ethanol attacks the carbon in the C=O of the carboxylic acid group producing a new bond between the oxygen in the ethanol and the carbon in the carboxylic acid. In <em>step 3</em>, a proton is transferred to produce a better leaving group (
). In <em>step 4</em>, a water molecule leaves the main structure to produce again the double bond C=O. <em>Finally</em>, a base (
) removes the hydrogen from the C=O bond to produce ethyl L-valinate
See figure 1
I hope it helps!
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