2H₂₍g₎ + O₂ ₍g₎→ 2H₂O
138 mol H₂ × (2 mol H₂O ÷ 2 mol H₂)= 138 mol H₂O
64 mol O₂ × (2 mol H₂O ÷ 1 mol O₂)= 128 mol H₂O
128 mol H₂O
I don’t see any equal signs to make it an equation. Am I missing something?
Answer: 2N2O3 -> 2N2 + 3O2
Explanation:
Answer:
= 200 mL
Explanation:
Using the dilution formula;
M1V1 = M2V2 ;
Where, M1 is the concentration before dilution, V2 is the volume before dilution, while M2 is the concentration after dilution and V2 is the volume after dilution.
M1 = 2.0 M
V1 = 50 mL
M2 = 0.50 M
V2 = ?
V2 = M1V1/M2
= ( 2.0 × 50 )/ 0.5
= 200 mL
Therefore, the volume after dilution will be, 200 mL
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!