<u>Answer:</u> The expression for
is written below.
<u>Explanation:</u>
Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of the products and the reactants each raised to the power their stoichiometric ratios. It is expressed as 
For a general chemical reaction:

The expression for
is written as:

The partial pressure for solids and liquids are taken as 1.
For the given chemical equation:

The expression for
for the following equation is:

The partial pressure of
will be 1 because it is solid.
So, the expression for
now becomes:

Hence, the expression for
is written above.
The heat of combustion for methanol is 727 kj/mol
<em><u>calculation</u></em>
calculate the moles of methanol (CH3OH)
moles = mass/molar mass
molar mass of methanol = 12 +( 1 x3) +16 + 1= 32 g /mol
moles is therefore= 64.0 g / 32 g/mol = 2 moles
Heat of combustion is therefore = 1454 Kj / 2 moles = 727 Kj/mol
Better than i am and very precice
You did not include the questions.
I did some research and found the questions:
<span>
What is the mass of 1 mole of pennies?
How many moles of pennies have a mass equal to the mass of the moon?
Solutions:
1) mass of 1 mole of pennies
Data: mass of 1 penny = 2.50 g
1 mole = 6.022 * 10^ 23 units
Proportion:
1 penny 6.022 * 10^23 penny
-------------- = ----------------------------
2.50 g x
Solve: x = 6.022 * 10^23 penny * 2.50g / 1 penny = 15.055* 10^23
Since 2.50 has 3 significant figures, the answer must use 3 significant figures => x = 15.1 * 10^ 23 g = 1.51 * 10^24 g
Answer: 1 mol of pennies have a mass of 1.51 * 10^24 g
2) How many moles of pennies have a mass equal to the same mass of the Moon
Convert the mass of the Moon grams: 7.35 * 10^22 kg = 7.35 * 10^ 25 g
1 mol x
---------------------- = ----------------------
1.51 * 10^ 24g 7.35 * 10^ 25 g
=> x = 7.35 * 10^ 25 g * 1 mol / (1.51 * 10^24 g)= 48.7 mol
Answer: 48.7 mol
</span>
Knowing the number of valence electrons in one of the alien elements helps in identifying it because the number of valence electrons can help categorize the alien element. Similar elements have the same valence electrons and knowing the category of the element can help further analyze the element.