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monitta
3 years ago
13

An emulsifying agent is typically characterized by having ____. a. one polar end c. two polar ends b. one nonpolar end d. one po

lar end and one nonpolar end
Chemistry
1 answer:
tangare [24]3 years ago
4 0

Answer:

An emulsifying agent is typically characterized by having <u><em>d. one polar end and one nonpolar end.</em></u>

Explanation:

Emulsifiers are substances that have the ability to bind, for example, fats with those substances that have mostly water in their conformation. In other words, the emulsifier facilitates mixtures of two or more immiscible liquid substances.

This is because the molecules of an emulsifier are often lipophilic (attract oil) at one end and hydrophilic (attract water) at the other. In other words it consists of a polar (hydrophilic) head group and a non-polar (hydrophobic) tail.

<u><em>An emulsifying agent is typically characterized by having d. one polar end and one nonpolar end.</em></u>

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A student titrates a 10.00mL sample of an HCl solution, using 0.359 M solution of NaOH. She finds that 24.75mL of sodium hydroxi
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HCl and NaOH react in a 1:1 ratio, meaning that 1 H+ from HCl will react with 1 OH- from NaOH. Knowing this, and that molarity is mol/liter, all we need to do is use what we have available. First we must find the mols of HCl in our solution, so we set up the following equation in the following steps:
1. 24.75mL x (0.359mol NaOH / 1000mL) = 8.885 x 10^-3mol NaOH
   This is done in order to find the mols of NaOH to convert to mols of HCl.
2. 8.885x10^-3mol NaOH x (1 mol HCl/1mol NaOH) = 8.885 x 10^-3mol HCl
   Here we just used the mols of NaOH we found to convert to mols of HCl using the 1:1 ratio described earlier.

From the mols of HCl all we have to do is divide by the amount of liters in the solution. Since we started with 10mL HCl and added 24.75mL NaOH, the total volume is 34.75mL = 0.03475L. So:
8.885 x 10^-3mol HCl/0.03475L = 2.557 x 10^-1M HCl
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M1 = the molarity of our HCl in the titrated mixture (2.557 x 10^-1M HCl)
V1 = the total volume that our mixture has (34.75mL = 0.03475L)
M2 = what we're trying to find
V2 = the amount of the original HCl that we had (10mL = 0.010L)
Simply solving for M2 gives us:
M2 = (M1V1) / V2 or:
M2=((2.557 x 10^-1) x 0.03475L) / 0.010L = 8.89 x 10^-1M HCl. That is your answer.
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