0.216g of aluminium compound X react with an excess of water water to produce gas. this gas burn completely in O2 to form H2O and 108cm^3of CO2 only . the volume of CO2 was measured at room temperature and pressure
0.108 / n = 24 / 1
n = 0.0045 mole ( CO2 >>0.0045 mole
0.216 - 0.0045 = 0.2115
so Al = 0.2115 / 27 => 0.0078 mole
C = 0.0045 * 1000 => 4.5 and Al = 0.0078 * 1000 = 7.8
The question is incomplete, complete question is :
The first two steps in the industrial synthesis of nitric acid produce nitrogen dioxide from ammonia:
Step 1 : 
Step 2 : 
The net reaction is:
Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants 
and 
. If you need to include any physical constants, be sure you use their standard symbols
Answer:
Equation that gives the overall equilibrium constant K in terms of the equilibrium constants 
:
Explanation:
Step 1 :
Expression of an equilibrium constant can be written as:
![K_1=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BNO%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E5%7D)
Step 2 :
Expression of an equilibrium constant can be written as:
![K_2=\frac{[NO_2]^2}{[NO]^2[O_2]}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BNO%5D%5E2%5BO_2%5D%7D)
The net reaction is:
Expression of an equilibrium constant can be written as:
![K=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO_2%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E7%7D)
Multiply and divide ![[NO]^4](https://tex.z-dn.net/?f=%5BNO%5D%5E4)
;
![K=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}\times \frac{[NO]^4}{[NO]^4}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO_2%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E7%7D%5Ctimes%20%5Cfrac%7B%5BNO%5D%5E4%7D%7B%5BNO%5D%5E4%7D)
![K=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^7}\times \frac{[NO_2]^4}{[NO]^4}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E7%7D%5Ctimes%20%5Cfrac%7B%5BNO_2%5D%5E4%7D%7B%5BNO%5D%5E4%7D)
![K=K_1\times \frac{[NO_2]^4}{[O_2]^2[NO]^4}](https://tex.z-dn.net/?f=K%3DK_1%5Ctimes%20%5Cfrac%7B%5BNO_2%5D%5E4%7D%7B%5BO_2%5D%5E2%5BNO%5D%5E4%7D)
![K=K_1\times (\frac{[NO_2]^2}{[O_2]^1[NO]^2})^2](https://tex.z-dn.net/?f=K%3DK_1%5Ctimes%20%28%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BO_2%5D%5E1%5BNO%5D%5E2%7D%29%5E2)
So , the equation that gives the overall equilibrium constant K in terms of the equilibrium constants :
Answer:
No.
Explanation: You cannot change the formulas of individual substances because the chemical formula for a given substance is characteristic of that substance.
It is a period.....
this ok??
Answer is: Kb for ammonia is 1.73·10⁻⁵.<span>
Chemical reaction of ammonia in water: NH</span>₃ + H₂O → NH₄⁺ +
OH⁻<span>.
Kb(NH</span>₃) = ?<span>.
c</span>₀(NH₃<span>) = 0.150 M.
c(NH</span>₄⁺) =
c(OH⁻<span>) = 0.0016 M.
c(NH</span>₃<span>) = 0.150 M - 0.0016.
</span>c(NH₃) = 0.1484 M.<span>
Kb</span>(NH₃) = c(NH₄⁺) · c(OH⁻) / c(NH₃<span>).
Kb</span>(NH₃) = (0.0016 M)² / 0.1484 M.
Kb(NH₃) = 1.73·10⁻⁵.<span>
</span>
