Answer:
An alcohol thermometer can measure the freezing point of a liquid that freezes at −80 °C.
Explanation:
A thermometer is a device used to measure temperature. A thermometer must contain a thermometric substance. A thermometric substance is any substance having a particular physical property that changes with temperature.
For all liquid-in-glass thermometers, the property that changes with change in temperature is the height of the liquid. There are two kinds of liquid-in-glass thermometers; mercury-in-glass thermometer and alcohol-in-glass thermometer.
Alcohol-in-glass thermometer measures very low temperatures up to as low as -115°C. If it measures such a low temperature, then it can efficiently measure -80°C hence the answer.
Alcohol-in-glass thermometers have a narrower temperature range than mercury-in-glass thermometer. The later is well adapter for the measurement bof higher tempetures up to 357°C.
<u>Answer:</u>
<em>Here the given material is taken and mixed with water.</em>
<u>Explanation:</u>
The amount of material and water taken are same. Hence if it is not soluble in water it should make a dense and flowy paste like material and if it is soluble in water it should this and thicker density of water should remain.
If the amount of water that we are taking is more than the material will float in water if it is not soluble and lighter than water or would sink if it is heavier than water.
Answer:
Initial concentration of HI is 5 mol/L.
The concentration of HI after 
is 0.00345 mol/L.
Explanation:
Rate Law: ![k[HI]^2 ](https://tex.z-dn.net/?f=k%5BHI%5D%5E2%0A)
Rate constant of the reaction = k = 
Order of the reaction = 2
Initial rate of reaction = 
Initial concentration of HI =![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
![1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E%7B-7%7D%20mol%2FL%20s%3D%286.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%29%5BHI%5D%5E2)
![[A_o]=5 mol/L](https://tex.z-dn.net/?f=%5BA_o%5D%3D5%20mol%2FL)
Final concentration of HI after t = [A]
t =
Integrated rate law for second order kinetics is given by:
![\frac{1}{[A]}=kt+\frac{1}{[A_o]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3Dkt%2B%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
![\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D6.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%5Ctimes%204.53%5Ctimes%2010%5E%7B10%7D%20s%2B%5Cfrac%7B1%7D%7B%5B5%20mol%2FL%5D%7D)
![[A]=0.00345 mol/L](https://tex.z-dn.net/?f=%5BA%5D%3D0.00345%20mol%2FL)
The concentration of HI after is 0.00345 mol/L.
B and D is out. It cant be A because heat of combustion is substance not compound. So the answer is D.
I would say the answer is emissions. These are the particles that are not supposed to be present in air but due to the production of different substances from humans daily activities these substances go with the air we breath. Hope this helped.
