Draw a heating curve for 1 mole of methanol beginning at 170K and ending at 350K. Assume that the values given here are constant
1 answer:
The graph of the heating of methanol has been attached.
The temperature at 170 K is approximately the melting point of methanol.
The temperature at 350 K is approximately the boiling point of methanol.
If we provide heat before 170 K temperature then all the heat provided to the methanol will result in rise in temperature of methanol in solid form.
If we provide heat at 170 K temperature then all the heat provided to the methanol will result in change in physical from of methanol from solid to liquid.
If we provide heat after 170 K temperature and before 350 K temperature then all the heat provided to the methanol will result in rise in temperature of methanol in liquid form.
If we provide heat at 340 K temperature then all the heat provided to the methanol will result in change in physical from of methanol from liquid to gas.
If we provide heat after 350 K temperature then all the heat provided to the methanol will result in rise in temperature of methanol in gas form.
So, the heat given to methanol will rise its temperature except at its Boiling and melting point. At those temperature heat will change the physical form of methanol first. and the graph also have been attached of between boiling point and melting point as given in question .
Learn more about methanol here:
#SPJ10
You might be interested in
Answer: -
92.4 s
Explanation: -
The decomposition reaction of a → b has a rate constant k = 0.00132 s⁻¹
From the rate constant we see that the reaction is of zero order.
The rate equation for a zero order reaction is
A₀ - A = kt
where A₀ = initial concentration.
T = time passed since start of reaction,
A is the amount present after t time passed.
A₀ = 0.156 M
A = A₀ - 78.1% of A₀
= 0.156 - x 0.156
= 0.156 - 0.122
= 0.034 M
Plugging into the formula
A₀ - A = kt
0.156 - 0.034 = 0.00132 x t
t =
= 92.4 s
t =