<span>Why are leaves different colors?</span><span>
The chlorophyll breaks down</span>
option (c) emulsion is the right answer
Emulsion is a classification of milk.
<h3>Describe an emulsion.</h3>
In physical chemistry, an emulsion is a combination of two or more liquids in which one of the liquids is present as microscopic or ultramicroscopic droplets dispersed throughout the other.
<h3>How is emulsion used?</h3>
Emulsions form as a result of the cleansing action of soaps.
(ii) The emulsification process is how lipids are broken down in the intestines.
(iii) Disinfectants and antiseptics combine with water to generate emulsions.
(iv) The emulsification technique is utilized to create medications.
<h3>How is an emulsion created?</h3>
When two immiscible liquids, such as oil and water, are stirred together with an emulsifier—which might be a protein, phospholipid, or even a nanoparticle—emulsion is created.
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Answer:
Here's what I get.
Explanation:
- If your teachers don't ask for a specific type of formula, a condensed structural formula should be OK.
- If they ask specifically for a structural formula or a bond-line formula, that is what you must give.
Bottom line: ask your teachers in advance what they expect.
The number of mole of HCl needed for the solution is 1.035×10¯³ mole
<h3>How to determine the pKa</h3>
We'll begin by calculating the pKa of the solution. This can be obtained as follow:
- Equilibrium constant (Ka) = 2.3×10¯⁵
- pKa =?
pKa = –Log Ka
pKa = –Log 2.3×10¯⁵
pKa = 4.64
<h3>How to determine the molarity of HCl </h3>
- pKa = 4.64
- pH = 6.5
- Molarity of salt [NaZ] = 0.5 M
- Molarity of HCl [HCl] =?
pH = pKa + Log[salt]/[acid]
6.5 = 4.64 + Log[0.5]/[HCl]
Collect like terms
6.5 – 4.64 = Log[0.5]/[HCl]
1.86 = Log[0.5]/[HCl]
Take the anti-log
0.5 / [HCl] = anti-log 1.86
0.5 / [HCl] = 72.44
Cross multiply
0.5 = [HCl] × 72.44
Divide both side by 72.44
[HCl] = 0.5 / 72.4
[HCl] = 0.0069 M
<h3>How to determine the mole of HCl </h3>
- Molarity of HCl = 0.0069 M
- Volume = 150 mL = 150 / 1000 = 0.15 L
Mole = Molarity x Volume
Mole of HCl = 0.0069 × 0.15
Mole of HCl = 1.035×10¯³ mole
<h3>Complete question</h3>
How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl
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