<h2>Selective & Differential Medium</h2>
Explanation:
- Selective media allow specific types of organisms to develop, and inhibit the development of different living beings. The selectivity is cultivated in a few ways.For model, living beings that can use a given sugar are handily screened by making that sugar the main carbon source in the medium. On the other hand,selective hindrance of certain sorts of microorganisms can be accomplished by adding dyes, anti-infection agents, salts or explicit inhibitors which influence the digestion or enzyme systems of the living beings
- Differential media are utilized to separate firmly related life forms or groups of living beings. owing to the pre of specific colors or synthetic compounds in the media, the creatures will deliver trademark changes or development designs that are utilized for ID or separation. An assortment of particular and differential media are utilized in clinical, demonstrative and water contamination research facilities, and in food and dairy laboratories
- Selective media because elevated NaCI level is designed to help grow selective bacteria.differential media because the fermented sugar gives off a yellow halo which allows for differentiate between bacteria
The control group in this experiment is the one with just distilled water. It is plain so they can use it to compare the other tests against.
Answer:
HCL and CaCl2
Explanation:
Since calcium hydroxide is a base, then it needs an acid to neutralise it.
HCL is an acid, it can neutralize calcium hydroxide through reacting the two together.
The resultant salt shall be calcium chloride ( CaCl2 )
Reaction of calcium hydroxide and HCl results into Ca(OH)Cl, CaCl2,
Where Ca(OH)Cl is an aqueous solution and calcium chloride is a base.
Hope the answer helps out.
<span>Some
of the solutions exhibit
colligative properties. These properties depend on the amount of solute
dissolved in a solvent. For boiling point elevation, we calculate the increase in temperature by the equation:
</span><span>ΔT(boiling point) =
(Kb)mi
where Kb is a constant, m is the molality of the solution, i is the van't Hoff factor.
From the given data, we can easily calculate for i as follows:
</span>ΔT(boiling point) = (Kb)mi
103.45 - 100 = (0.512)3.90i
i = 1.73 <-------van't Hoff factor
