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: The orange color remains unchanged. (B)
Explanation:
Answer:
The temperature associated with this radiation is 0.014K.
Explanation:
If we assume that the astronomical object behaves as a black body, the relation between its <em>wavelength</em> and <em>temperature</em> is given by Wien's displacement law.
where,
λmax is the wavelength at the peak of emission
b is Wien's displacement constant (2.89×10⁻³ m⋅K)
T is the absolute temperature
For a wavelength of 21 cm,
B
When frequency increases, as does the energy, but wavelength decreases. It also works vise versa; if wavelength were to increase, its frequency and energy will decrease.
