"Pent" is five, "ene" means double bonded carbon
Answer:
First, place no. 5 in front of the CO2 in order to balance the carbon atoms. Next, place no. 6 in front of H2O to balance the hydrogen atoms. Lastly place no. 8 in front of the O2 so that there are 16 oxygen atoms on both sides of the reaction.
I'm not 100% sure but I'm leaning towards D. :)
Answer:
Explanation:
2 mole x 6.023x 10^23atoms/mole =12.04 atoms
Answer:
24x10³
Explanation:
2CO₂(g) + 4H₂O(g) → 2CH₃OH(l) + 3O₂ (g)
The equilibrium constant for this reaction is:
Kc = ![\frac{[O_2]^3}{[CO_2]^2[H_2O]^4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BO_2%5D%5E3%7D%7B%5BCO_2%5D%5E2%5BH_2O%5D%5E4%7D)
The expression of [CH₃OH] is left out as it is a pure liquid.
Now we <u>convert the given masses of the relevant species into moles</u>, using their <em>respective molar masses</em>:
- CO₂ ⇒ 3.28 g ÷ 44 g/mol = 0.0745 mol CO₂
- H₂O ⇒ 3.86 g ÷ 18 g/mol = 0.214 mol H₂O
- O₂ ⇒ 2.80 g ÷ 32 g/mol = 0.0875 mol O₂
Then we calculate the concentrations:
- [CO₂] = 0.0745 mol / 7.5 L = 0.0099 M
- [H₂O] = 0.214 mol / 7.5 L = 0.0285 M
- [O₂] = 0.0875 mol / 7.5 L = 0.0117 M
Finally we <u>calculate Kc</u>:
- Kc =
= 24x10³