Answer : The standard enthalpy of formation of methanol is, -238.7 kJ/mole
Explanation :
Standard formation of reaction : It is a chemical reaction that forms one mole of a substance from its constituent elements in their standard states.
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The formation reaction of
will be,

The intermediate balanced chemical reaction will be,
(1)

(2)

(3)

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, we get :
(1)

(2)

(3)

The expression for enthalpy of formation of
will be,



Therefore, the standard enthalpy of formation of methanol is, -238.7 kJ/mole
The correct option is NONE OF THE ABOVE. T
The bases in the options above such as NaOH and KOH are strong hydroxides and they are strong bases. Examples of hydroxides that are weak bases are iron hydroxide, ammonium hydroxide, etc.
<h3>
Answer:</h3>
0.95 atm
<h3>
Explanation:</h3>
We are given;
Initial pressure, P1 = 1.0 atm
Initial temperature, T1 =298 K (25°C + 273)
Initial volume, V1 = 0.985 L
Final temperature, T2 = 295 K (22°C + 273)
Final volume, V2 = 1.030 L
We are required to find final air pressure;
Using the combined gas law;

To get, P2 ;



= 0.95 atm
Therefore, the air pressure at the top of the mountain is 0.95 atm
Because energy cannot be created or destroyed, only transferred