Explanation:

Equilibrium constant of reaction = 
Concentration of NO = ![[NO]=\frac{2.69\times 10^{-2} mol}{1 L}=2.69\times 10^{-2} M](https://tex.z-dn.net/?f=%5BNO%5D%3D%5Cfrac%7B2.69%5Ctimes%2010%5E%7B-2%7D%20mol%7D%7B1%20L%7D%3D2.69%5Ctimes%2010%5E%7B-2%7D%20M)
Concentration of bromine gas = ![[Br_2]=\frac{3.85\times 10^{-2} mol}{1 L}=3.85\times 10^{-2} M](https://tex.z-dn.net/?f=%5BBr_2%5D%3D%5Cfrac%7B3.85%5Ctimes%2010%5E%7B-2%7D%20mol%7D%7B1%20L%7D%3D3.85%5Ctimes%2010%5E%7B-2%7D%20M)
Concentration of NOBr gas = ![[Br_2]=\frac{9.56\times 10^{-2} mol}{1 L}=9.56\times 10^{-2} M](https://tex.z-dn.net/?f=%5BBr_2%5D%3D%5Cfrac%7B9.56%5Ctimes%2010%5E%7B-2%7D%20mol%7D%7B1%20L%7D%3D9.56%5Ctimes%2010%5E%7B-2%7D%20M)
The reaction quotient is given as:
![Q=\frac{[NOBr]^2}{[NO]^2[Br_2]}=\frac{(9.56\times 10^{-2} M)^2}{(2.69\times 10^{-2} M)^2\times 3.85\times 10^{-2} M}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BNOBr%5D%5E2%7D%7B%5BNO%5D%5E2%5BBr_2%5D%7D%3D%5Cfrac%7B%289.56%5Ctimes%2010%5E%7B-2%7D%20M%29%5E2%7D%7B%282.69%5Ctimes%2010%5E%7B-2%7D%20M%29%5E2%5Ctimes%203.85%5Ctimes%2010%5E%7B-2%7D%20M%7D)


The reaction will go in backward direction in order to achieve an equilibrium state.
1. In order to reach equilibrium NOBr (g) must be produced. False
2. In order to reach equilibrium K must decrease. False
3. In order to reach equilibrium NO must be produced. True
4. Q. is less than K . False
5. The reaction is at equilibrium. No further reaction will occur. False
The ________ orbital is degenerate with 5py in a many-electron atom.
<h2>
5px is the correct answer</h2>
Answer:
B) 7.7
Explanation:
For the reaction Ag2CO3(s) + CrO42‒(aq) → Ag2CrO4(s) + CO32‒(aq)
Kc = (CO₃²⁻) / (CrO₄²⁻)
and the Ksp given are
Ag₂CO₃ ⇒ 2 Ag⁺(aq) + CO₃²⁻(aq) Ksp₁ = (Ag⁺)²(CO₃²⁻)
Ag₂CrO₄ ⇒ 2 Ag⁺(aq)+ CrO₄²⁻(aq) Ksp₂ = (Ag⁺)²(CrO₄²⁻)
Where (...) indicate concentrations M
Notice if we divide the expressions for Ksp we get:
Ksp₁/Ksp₂ = (CO₃²⁻) / (CrO₄²⁻) = 8.5 x 10⁻¹² / 1.1 x 10⁻¹² = 7.7
which is the desired answer.
Answer:

Explanation:
1. Write the skeleton equation for the half-reaction
NO₃⁻ ⟶ N₂O
2. Balance all atoms other than H and O
2NO₃⁻ ⟶ N₂O
3. Balance O by adding H₂O molecules to the deficient side.
2NO₃⁻ ⟶ N₂O + 5H₂O
4. Balance H by adding H⁺ ions to the deficient side.
2NO₃⁻ + 10H⁺ ⟶ N₂O + 5H₂O
5. Balance charge by adding electrons to the deficient side.
2NO₃⁻ + 10H⁺ + 8e⁻ ⟶ N₂O + 5H₂O
The amount of charge required to reduce 2 mol of NO₃⁻ is 8 F
