Answer : The normal boiling point of ethanol will be,
or ![75.67^oC](https://tex.z-dn.net/?f=75.67%5EoC)
Explanation :
The Clausius- Clapeyron equation is :
![\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})](https://tex.z-dn.net/?f=%5Cln%20%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7BR%7D%5Ctimes%20%28%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%29)
where,
= vapor pressure of ethanol at
= 98.5 mmHg
= vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg
= temperature of ethanol = ![30^oC=273+30=303K](https://tex.z-dn.net/?f=30%5EoC%3D273%2B30%3D303K)
= normal boiling point of ethanol = ?
= heat of vaporization = 39.3 kJ/mole = 39300 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
![\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})](https://tex.z-dn.net/?f=%5Cln%20%28%5Cfrac%7B760mmHg%7D%7B98.5mmHg%7D%29%3D%5Cfrac%7B39300J%2Fmole%7D%7B8.314J%2FK.mole%7D%5Ctimes%20%28%5Cfrac%7B1%7D%7B303K%7D-%5Cfrac%7B1%7D%7BT_2%7D%29)
![T_2=348.67K=348.67-273=75.67^oC](https://tex.z-dn.net/?f=T_2%3D348.67K%3D348.67-273%3D75.67%5EoC)
Hence, the normal boiling point of ethanol will be,
or ![75.67^oC](https://tex.z-dn.net/?f=75.67%5EoC)
Answer:
growth limit for trees
Explanation:
the awnser is the growth limit for trees
Sodium is considered to be a transitional metal
Moss leaves and fish are different in that, the moss leave is a producer, that is, it produces its own food through photosynthesis while the fish is a consumer, it feeds on foods that are not produced by it.
Both moss and fish are the same in the sense that both have cell as their basic unit of life, that is, they both possess cells.
Answer : The metal used was iron (the specific heat capacity is
).
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
![q_1=-q_2](https://tex.z-dn.net/?f=q_1%3D-q_2)
![m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_f-T_1%29%3D-m_2%5Ctimes%20c_2%5Ctimes%20%28T_f-T_2%29)
where,
= specific heat of unknown metal = ?
= specific heat of water = ![4.184J/g^oC](https://tex.z-dn.net/?f=4.184J%2Fg%5EoC)
= mass of unknown metal = 150 g
= mass of water = 200 g
= final temperature of water = ![34.3^oC](https://tex.z-dn.net/?f=34.3%5EoC)
= initial temperature of unknown metal = ![150.0^oC](https://tex.z-dn.net/?f=150.0%5EoC)
= initial temperature of water = ![25.0^oC](https://tex.z-dn.net/?f=25.0%5EoC)
Now put all the given values in the above formula, we get
![150g\times c_1\times (34.3-150.0)^oC=-200g\times 4.184J/g^oC\times (34.3-25.0)^oC](https://tex.z-dn.net/?f=150g%5Ctimes%20c_1%5Ctimes%20%2834.3-150.0%29%5EoC%3D-200g%5Ctimes%204.184J%2Fg%5EoC%5Ctimes%20%2834.3-25.0%29%5EoC)
![c_1=0.449J/g^oC](https://tex.z-dn.net/?f=c_1%3D0.449J%2Fg%5EoC)
Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).
Therefore, the metal used was iron (the specific heat capacity is
).