Number of molecules in 1 dm³ Oxygen = 2.71 x 10²²
<h3>Further explanation</h3>
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters/mol.
The mole is the number of particles contained in a substance
1 mol = 6.02.10²³
1 dm³ of oxygen = 1 L Oxygen

n=mol=0.045
No = 6.02.10²³

3. A
4. B
5. A
6. E
7. A
8. C
9. A
10. B
Some of these were guesses but they were educated guesses. Best of luck. If some of them are wrong I am sorry. <span />
Is there something else that goes along with this?
Answer:blue blue blue
Explanation:
Just answered it and got it right
<u>Answer:</u> The equilibrium constant for the given reaction is 0.8
<u>Explanation:</u>
Equilibrium constant is defined as the ratio of concentration of the products raised to the power its stoichiometric coefficients to the concentration of reactants raised to power its stoichiometric coefficient. It is represented as 
For the general equation:

The equilibrium constant is represented as:
![K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
For the given chemical equation:

for this equation is given by:
![K_c=\frac{[H_2O][CO]}{[H_2][CO_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2O%5D%5BCO%5D%7D%7B%5BH_2%5D%5BCO_2%5D%7D)
Concentration at equilibrium of

Putting values in above equation, we get:

Hence, the equilibrium constant for the given chemical reaction is 0.8