Answer:
![K=K_1*K_2\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}](https://tex.z-dn.net/?f=K%3DK_1%2AK_2%5C%5C%5C%5CK%3D%5Cfrac%7B%5BH_2%5D%5E3%5BCO_2%5D%5BH_2%5D%7D%7B%5BCH_4%5D%5BH_2O%5D%5BH_2O%5D%7D)
Explanation:
Hello there!
In this case, for the given chemical reaction, it turns out firstly necessary to write the equilibrium expression for both reactions 1 and 2:
![K_1=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \\\\K_2=\frac{[CO_2][H_2]}{[CO][H_2O]}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BCO%5D%5BH_2%5D%5E3%7D%7B%5BCH_4%5D%5BH_2O%5D%7D%20%5C%5C%5C%5CK_2%3D%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D)
Now, when we combine them to get the overall expression, we infer these two are multiplied to get:
![K=K_1*K_2\\\\K=\frac{[CO][H_2]^3}{[CH_4][H_2O]} *\frac{[CO_2][H_2]}{[CO][H_2O]}\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}](https://tex.z-dn.net/?f=K%3DK_1%2AK_2%5C%5C%5C%5CK%3D%5Cfrac%7B%5BCO%5D%5BH_2%5D%5E3%7D%7B%5BCH_4%5D%5BH_2O%5D%7D%20%2A%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D%5C%5C%5C%5CK%3D%5Cfrac%7B%5BH_2%5D%5E3%5BCO_2%5D%5BH_2%5D%7D%7B%5BCH_4%5D%5BH_2O%5D%5BH_2O%5D%7D)
Regards!
Answer:
123.2 Liters.
Explanation:
At STP (T = 273K & P = 1atm)<em>, one mol of any gas will occupy 22.4 liters</em>.
With the above information in mind, we can <u>calculate how many liters would 5.500 mol of gas occupy</u>:
5.500 mol * 22.4 L / mol = 123.2 L
So 5.500 moles of C₃H₃ would have a volume of 123.2 liters at STP.
The question is incomplete, here is the complete question:
A bottle of rubbing alcohol having aqueous solution of alcohol contains 70% (v/v) alcohol. If Carl buys a 946 ml bottle of rubbing alcohol, how much of the aqueous solution is water?
<u>Answer:</u> The amount of water present in the given bottle of rubbing alcohol is 283.8 mL
<u>Explanation:</u>
We are given:
Volume of bottle of rubbing alcohol = 946 mL
70% (v/v) alcohol solution
This means that 70 mL of rubbing alcohol is present in 100 mL of solution
Amount of water present in solution = [100 - 70] = 30 mL
Applying unitary method:
In 100 mL of solution, the amount of water present is 30 mL
So, in 946 mL of solution, the amount of water present will be = 
Hence, the amount of water present in the given bottle of rubbing alcohol is 283.8 mL