Answer:
The equilibrium constant Ksp of the generic salt AB2 = 6.4777 *10^-8 M
Explanation:
Step 1: The balanced equation
AB2 ⇒ A2+ + 2B-
Step 2: Given data
Concentration of A2+ = 0.00253 M
Concentration of B- = 0.00506 M
Step 3: Calculate the equilibrium constant
Equilibrium constant Ksp of [AB2] = [A2+][B-]²
Ksp = 0.00253 * 0.00506² = 6.4777 *10^-8 M
The equilibrium constant Ksp of the generic salt AB2 = 6.4777 *10^-8 M
Answer:
0.42 M
Explanation:
The reaction that takes place is:
- Cu(CH₃COO)₂ + Na₂CrO₄ → Cu(CrO₄) + 2Na(CH₃COO)
First we <u>calculate the moles of Na₂CrO₄</u>, using the <em>given volume and concentration</em>:
(200 mL = 0.200L)
- 0.70 M * 0.200 L = 0.14 moles Na₂CrO₄
Now we <u>calculate the moles of Cu(CH₃COO)₂</u>, using its <em>molar mass</em>:
- 40.8 g ÷ 181.63 g/mol = 0.224 mol Cu(CH₃COO)₂
Because the molar ratio of Cu(CH₃COO)₂ and Na₂CrO₄ is 1:1, we can directly <u>substract the reacting moles of Na₂CrO₄ from the added moles of Cu(CH₃COO)₂</u>:
- 0.224 mol - 0.14 mol = 0.085 mol
Finally we <u>calculate the resulting molarity</u> of Cu⁺², from the <em>excess </em>cations remaining:
- 0.085 mol / 0.200 L = 0.42 M
Since chromium has 24 as its atomic number (which mean that it has 24 proton and 24 electron in neutral ground state), the electron configuration for the chromium would be :
1s^2 , 2s^2 , 2p^6 , 3s^2, 3p^6 , 3d^5 , 4s^1
Answer:
pH = 4.45
Explanation:
We need to find the pH of 
solution of HCl. We know that, pH of a solution is given by :
![pH=-log[H]^+](https://tex.z-dn.net/?f=pH%3D-log%5BH%5D%5E%2B)
Put all the values,
![pH=-log[3.5\times 10^{-5}]\\\\pH=4.45](https://tex.z-dn.net/?f=pH%3D-log%5B3.5%5Ctimes%2010%5E%7B-5%7D%5D%5C%5C%5C%5CpH%3D4.45)
So, the pH of the solution of HCl is 4.45.
Answer:1
Explanation: There is only one orbital in the s sublevel. It can hold up to two electrons.
