Three examples of environmental, industrial and bio-chemistry are listed below:
- Environmental chemistry: Contamination, Atmospheric Deposition, and Soil Pollution.
- industrial chemistry: industrial inorganic chemicals, industrial organic chemicals, and agricultural chemicals
- bio-chemistry: genetic, immunology, and enzymology
<h3>Meaning of Chemistry</h3>
Chemistry can be defined as a branch of science which is concerned with the substances matter is composed of, their properties and reactions,
Chemistry also deals with the use of such reactions to form new substances.
In conclusion, Three examples of environmental, industrial and bio-chemistry are listed anove
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The Molar concentration of your analyte solution is 1.17 m
<h3>What is titration reaction?</h3>
- Titration is a chemical analysis procedure that determines the amount of a sample's ingredient by adding a precisely known amount of another substance to the measured sample, with which the desired constituent reacts in a specific, known proportion.
Make use of the titration formula.
The formula is molarity (M) of the acid x volume (V) of the acid = molarity (M) of the base x volume (V) of the base.
if the titrant and analyte have a 1:1 mole ratio. (Molarity is a measure of a solution's concentration represented as the number of moles of solute per litre of solution.)
26 x 1.8 = 40 x M
M = 26 x1.8 /40
M = 1.17
The Molar concentration of your analyte solution is 1.17 m
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<u>Answer:</u> The molality of the solution is 0.1 m.
<u>Explanation:</u>
To calculate the molality of solution, we use the equation:
Where,
= Given mass of solute = 27.1 g
= Molar mass of solute = 27.1 g/mol
= Mass of solvent = 100 g
Putting values in above equation, we get:
Hence, the molality of the solution is 0.1 m.
Answer:
Titrations. Because a noticeable pH change occurs near the equivalence point of acid-base titrations, an indicator can be used to signal the end of a titration. When selecting an indicator for acid-base titrations, choose an indicator whose pH range falls within the pH change of the reaction.
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Answer:
One extraction: 50%
Two extractions: 75%
Three extractions: 87.5%
Four extractions: 93.75%
Explanation:
The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.
qⁿ = (V₁/(V₁ + KV₂))ⁿ
We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:
qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ = (V₁/(2V₁))ⁿ = (1/2)ⁿ
When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.
When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.
When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.
When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.