Tincture of iodine has antiseptic properties and is made by dissolving iodine crystals in alcohol.
<h3>What are antiseptics?</h3>
Antiseptics are substance which prevent or stops the growth of microorganisms.
Microorganisms have been known to cause diseases, therefore the application of antiseptics are useful to prevent diseases.
An example of an antiseptic is tincture of iodine.
Tincture of iodine is prepared by dissolving iodine crystals in alcohol.
In conclusion, antiseptics are useful in the prevention of diseases by microorganism. An example of an antiseptic is tincture of iodine.
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Answer:C
Explanation:The mineral is only formed when hot water cooler
<u>Answer:</u> The correct answer is Option 4.
<u>Explanation:</u>
Bromothymol blue, Bromocresol green and Thymol blue are the indicators which change their color according to the change in pH of the solution.
The pH range and color change of these indicators are:
- Bromothymol Blue: The pH range for this indicator is 6.0 to 7.5 and color change is from yellow to blue. It appears yellow below pH 6.0 and blue above pH 7.5
- Bromocresol green: The pH range for this indicator is 3.5 to 6.0 and color change is from yellow to blue. It appears yellow below pH 3.5 and blue above pH 6.0
- Thymol Blue: The pH range for this indicator is 8.0 to 9.6 and color change is from yellow to blue. It appears yellow below pH 8.0 and blue above pH 9.6
As, the highest pH of all the indicators is 9.6, so every indicator will appear blue above pH 9.6.
Hence, the correct answer is Option 4.
<u>Answer:</u> The value of for the net reaction is
<u>Explanation:</u>
The given chemical equations follows:
<u>Equation 1:</u>
<u>Equation 2:</u>
The net equation follows:
As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.
The value of equilibrium constant for net reaction is:
We are given:
Putting values in above equation, we get:
Hence, the value of for the net reaction is