Answer:
The heat of vaporisation of methanol is "3.48 KJ/Mol"
Explanation:
The amount of heat energy required to convert or transform 1 gram of liquid to vapour is called heat of vaporisation
When 8.7 KJ of heat energy is required to vaporize 2.5 mol of liquid methanol.
Hence, for 1 mol of liquid methanol, amount of heat energy required to evaporate the methanol is = 
= 3.48 KJ
So, the heat of vaporization 
Therefore, the heat of vaporization of methanol is 3.48KJ/Mol
Answer:
1,300,000,000,000
Explanation:
1.3 x 10^12
We want to convert this from scientific notation.
Tip: in scientific notation the exponent tells you how many place you move the decimal point over. If the exponent is negative you move the decimal point to the left. Ex. For, 4.1 x 10^-8, we would move the decimal point over 8 times to the left to get .00000041. When the exponent is positive we move over to the right. Ex. For, 7.6 x 10^7 we would move the decimal point over 7 times to the right to get 76,000,000
So to convert 1.3 x 10^12 we simply move over the decimal point over 12 times to the right.
1.3 x 10^12 ------> 1,300,000,000,000
Our answer is 1,300,000,000,000
Nitrogen and oxygen are in unpolluted air