Answer : The correct option is, (3) change states of matter.
Explanation :
Latent heat : It is defined as the heat required to convert the solid into liquid or vapor and a liquid into a vapor without changing the temperature.
There are two types of latent heat.
(1) Latent heat of fusion
(2) Latent heat of vaporization
Latent heat of fusion : It is defined as the amount of heat energy released or absorbed when the solid converted to liquid at atmospheric pressure at its melting point.
Latent heat of vaporization : It is defined as the amount of heat energy released or absorbed when the liquid converted to vapor at atmospheric pressure at its boiling point.
Hence, latent heat is used to change states of matter.
Answer:
25.907°C
Explanation:
In Exercise 102, heat capacity of bomb calorimeter is 6.660 kJ/°C
The heat of combustion of benzoic acid is equivalent to the total heat energy released to the bomb calorimeter and water in the calorimeter.
Thus:
= heat of combustion of benzoic acid
= heat energy released to water
= heat energy released to the calorimeter
Therefore,
![-m_{combust}*H_{combust} = [m_{water}*c_{water} + C_{calori}]*(T_{f} - T_{i})](https://tex.z-dn.net/?f=-m_%7Bcombust%7D%2AH_%7Bcombust%7D%20%3D%20%5Bm_%7Bwater%7D%2Ac_%7Bwater%7D%20%2B%20C_%7Bcalori%7D%5D%2A%28T_%7Bf%7D%20-%20T_%7Bi%7D%29)
1.056*26.42 = [0.987*4.18 + 6.66](
- 23.32)
27.8995 = [4.12566+6.660]( - 23.32)
( - 23.32) = 27.8995/10.7857 = 2.587
= 23.32 + 2.587 = 25.907°C
The protons! Is it the answer
In order to calculate the mass of nitrogen, we must first calculate the mass percentage of nitrogen in potassium nitrate. This is:
% nitrogen = mass of nitrogen / mass of potassium nitrate
% nitrogen = 14 / 101.1 x 100
The mass of nitrogen = % nitrogen x sample mass
= (14 / 101.1) x 101.1
= 14 grams
The molar weight of nitrogen is 14. Each mole of urea contains two moles of nitrogen. Therefore, for there to be 14 grams of nitrogen, there must be 0.5 moles of urea.
Mass of urea = moles urea x molecular weight urea
Mass of urea = 0.5 x 66.06
Mass of urea = 33.03 grams
Following are the possible isomers of secondary alcohol and ketones for six carbon molecules. In order to distinguish between sec. alcohol and ketone we can simply treat the unknown compound with acidified Potassium Dichromate (VI) in the presence of acid. If with treatment with unknown compound the colour of K2Cr2O7 (potassium dichromate VI) changes from orange to green then it is confirmed that the unknown compound is sec. alcohol, or if no change in colour is detected then ketone is confirmed. This is because ketone can not be further oxidized while, sec. alcohol can be oxidized to ketones as shown below,
