<u>Answer:</u> The concentration of ethyl ethanoate at equilibrium is 
<u>Explanation:</u>
Given values:
Equilibrium concentration of ethanol = 
Equilibrium concentration of ethanoic acid = 

The given chemical equation follows:

The expression of
for above equation follows:
![K_c=\frac{[C_2H_5OH][CH_3COOH]}{[CH_3COOC_2H_5]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC_2H_5OH%5D%5BCH_3COOH%5D%7D%7B%5BCH_3COOC_2H_5%5D%7D)
Putting values in above expression, we get:
![0.27=\frac{0.42\times 0.42}{[CH_3COOC_2H_5]}](https://tex.z-dn.net/?f=0.27%3D%5Cfrac%7B0.42%5Ctimes%200.42%7D%7B%5BCH_3COOC_2H_5%5D%7D)
![[CH_3COOC_2H_5]=\frac{0.42\times 0.42}{0.27}=0.653mol/dm^3](https://tex.z-dn.net/?f=%5BCH_3COOC_2H_5%5D%3D%5Cfrac%7B0.42%5Ctimes%200.42%7D%7B0.27%7D%3D0.653mol%2Fdm%5E3)
Hence, the concentration of ethyl ethanoate at equilibrium is 
Answer:
1. Find the molar mass of all the elements in the compound in grams per mole.
2. Find the molecular mass of the entire compound.
3. Divide the component's molar mass by the entire molecular mass.
4. You will now have a number between 0 and 1. Multiply it by 100% to get percent composition.
Explanation:
I think this one is the “a wall of water that is pushed ashore”