Remark
The short Answer is 6. That's why the ion has a charge of minus 2. Oxygen is doing it's best to have its outer ring have 8 electrons which is the number of outer electrons contained in the outer ring of the Noble Gas Neon.
Answer 6.
Answer:
Assignment: 06.05 Data Observation Assignment: 06.05 Data Observation Assignment: 06.05 Data Observation
Explanation:
Assignment: 06.05 Data Observation Assignment: 06.05 Data Observation Assignment: 06.05 Data Observation Assignment: 06.05 Data Observation Assignment: 06.05 Data Observation Assignment: 06.05 Data Observation Assignment: 06.05 Data Observation Assignment: 06.05 Data Observation Assignment: 06.05 Data Observation
Equation is as follow,
<span> 2 AgNO</span>₃<span> + MgBr</span>₂<span> </span>→ <span>2 AgBr + Mg(NO</span>₃<span>)</span>₂
According to eq.
339.74 g (2 moles) AgNO₃ produces = 375.54 g (2 moles) of AgBr
So,
22.5 g AgNO₃ will produce = X g of AgBr
Solving for X,
X = (22.5 g × 375.54 g) ÷ 339.74 g
X = 24.87 g of AgBr
Answer:
Explanation:
Step 1: Data given
The equilibrium constant, Kc, for the following reaction is 4.76 * 10^-4 at 431 K
The equilibrium concentration of Cl2(g) is 0.233 M
Step 2: The balanced equation
PCl5(g) ⇄ PCl3(g) + Cl2(g)
Step 3: The initial concentration
[PCl5]= Y M
[PCl] = 0M
[Cl2] = 0M
Step 4: Calculate the concentration at equilibrium
[PCl5] = Y + X M = Y - 0.233 M
[PCl]= XM = 0.233 M
[Cl2]= XM = 0.233 M
Step 5: Define Kc
Kc = [Cl2]* [PCl3] / [PCl5]
4.76 * 10^-4 = 0.233² / (Y -0.233)
0.000476 = 0.05429 / (Y - 0.233)
Y - 0.233 = 0.05429 / 0.000476
Y - 0.233 = 114.05 M
Y = 114.283 M = the initial concentration
The concentration of PCl5 at the equilibrium is 114.05 M
