Answer:
Kp = \frac{P(NH_{3}) ^{4} P(O_{2}) ^{5}}{P(NO) ^{4} P(H_{2}O)^{6}}
Explanation:
First, we have to write the balanced chemical equation for the reaction. Nitrogen monoxide (NO) reacts with water (H₂O) to give ammonia (NH₃) and oxygen (O₂), according to the following:
NO(g) + H₂O(g) → NH₃(g) + O₂(g)
To balance the equation, we add the stoichiometric coefficients (4 for NH₃ and NO to balance N atoms, then 6 for H₂O to balance H atoms and then 5 for O₂ to balance O atoms):
4 NO(g) + 6 H₂O(g) → 4 NH₃(g) + 5 O₂(g)
All reactants and products are in the gaseous phase, so the equilibrium constant is expressed in terms of partial pressures (P) and is denoted as Kp. The Kp is expressed as the product of the reaction products (NH₃ and O₃) raised by their stoichiometric coefficients (4 and 5, respectively) divided into the product of the reaction reagents (NO and H₂O) raised by their stoichiometric coefficients (4 and 6, respectively). So, the pressure equilibrium constant expression is written as follows:

<span>So when the chemist combines Ethane (CH3CH3) and Chlorine (Cl2) with the intention of producing Chloroethane (CH3CH2Cl), the other product that's formed in this reaction is 1,2-dichloroethane (ClCH2CH2Cl) also called as Ethylene dichloride with molecular weight of 98.954 g/mol. This is a colorless oily flammable substance that weighs heaver when vaporized.</span>
The percent yield of the reaction : 89.14%
<h3>Further explanation</h3>
Reaction of Ammonia and Oxygen in a lab :
<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>
mass NH₃ = 80 g
mol NH₃ (MW=17 g/mol):

mass O₂ = 120 g
mol O₂(MW=32 g/mol) :

Mol ratio of reactants(to find limiting reatants) :

mol of H₂O based on O₂ as limiting reactants :
mol H₂O :

mass H₂O :
4.5 x 18 g/mol = 81 g
The percent yield :

Answer:
pOH = 4.8
pH = 9.2
Explanation:
Given data:
Hydrogen ion concentration = 6.3×10⁻¹⁰M
pH of solution = ?
pOH of solution = ?
Solution:
Formula:
pH = -log [H⁺]
[H⁺] = Hydrogen ion concentration
We will put the values in formula to calculate the pH.
pH = -log [6.3×10⁻¹⁰]
pH = 9.2
To calculate the pOH:
pH + pOH = 14
We will rearrange this equation.
pOH = 14 - pH
now we will put the values of pH.
pOH = 14 - 9.2
pOH = 4.8