Answer: Conductor
Explanation: A material that conducts thermal energy well is called a thermal conductor.
Answer:
The desirable characteristic of Barium suspension are:
<u>Thick layering</u>
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Explanation:
First know the meaning of the terms :
1. Rapid flow
2. Thick layering
3. Good mucosal adhesion\
1. Rapid flow : The lower the viscosity of the solution , more is its flow in the solution. For example : Honey has more viscosity so it flows slowly . On the other hand water flow faster.
2. Thick layering : layering is the deposition of one layer of the liquid over other. The deposited layer can be of different substance.
3. Mucosal Adhesion : It is the adhesion(attraction) between two materials ,one of which is mucous.It is attractive force between the drug and the mucus inside the body.
Barium suspension:
<em>It is not a Good mucosal adhesion material because of low solubility in water and lipid .</em>
<em>Only those solution are good in adhesion which have higher solubility in water.</em>
It is widely used as the contrast media for the examination of gastrointestinal because of its <u>Thick Layering Property. It has the consistency of thick glass . At room temperature it is called "warm thick milk"</u>
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Answer: C.
Explanation: Alcohol floats on oil and water sinks in oil. Water, alcohol, and oil layer well because of their densities, but also because the oil layer does not dissolve in either liquid. The oil keeps the water and alcohol separated so that they do not dissolve in one another. ... Water sinks because it is more dense than oil.
Answer:
a)4.51
b) 9.96
Explanation:
Given:
NaOH = 0.112M
H2S03 = 0.112 M
V = 60 ml
H2S03 pKa1= 1.857
pKa2 = 7.172
a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.
Therefore, the half points will also be the middle point.
Solving, we have:
pH = (½)* pKa1 + pKa2
pH = (½) * (1.857 + 7.172)
= 4.51
Thus, pH at first equivalence point is 4.51
b) pH at second equivalence point:
We already know there is a presence of SO3-2, and it ionizes to form
SO3-2 + H2O <>HSO3- + OH-
![Kb = \frac{[ HSO3-][0H-]}{SO3-2}](https://tex.z-dn.net/?f=%20Kb%20%3D%20%5Cfrac%7B%5B%20HSO3-%5D%5B0H-%5D%7D%7BSO3-2%7D)
[HSO3-] = x = [OH-]
mmol of SO3-2 = MV
= 0.112 * 60 = 6.72
We need to find the V of NaOh,
V of NaOh = (2 * mmol)/M
= (2 * 6.72)/0.122
= 120ml
For total V in equivalence point, we have:
60ml + 120ml = 180ml
[S03-2] = 6.72/120
= 0.056 M
Substituting for values gotten in the equation ![Kb=\frac{[HSO3-][OH-]}{[SO3-2]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BHSO3-%5D%5BOH-%5D%7D%7B%5BSO3-2%5D%7D%20)
We noe have:
![x = [OH-] = 9.11*10^-^5](https://tex.z-dn.net/?f=x%20%3D%20%5BOH-%5D%20%3D%209.11%2A10%5E-%5E5)
=4.04
pH = 14- pOH
= 14 - 4.04
= 9.96
The pH at second equivalence point is 9.96
You can use a fan on the wire. That's one way.
