Answer:
0.3023 M
Explanation:
Let Picric acid = 
So,
+
⇄
+ 
The ICE table can be given as:
+
⇄
+ 
Initial: 0.52 0 0
Change: - x + x + x
Equilibrium: 0.52 - x + x + x
Given that;
acid dissociation constant (
) = 0.42
![K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}](https://tex.z-dn.net/?f=K_a%20%3D%20%5Cfrac%7B%5BH_3O%5E%2B%5D%5BPicric%5E-%5D%7D%7BH_%7Bpicric%7D%7D)
![0.42 = \frac{[x][x]}{0.52-x}}](https://tex.z-dn.net/?f=0.42%20%3D%20%5Cfrac%7B%5Bx%5D%5Bx%5D%7D%7B0.52-x%7D%7D)
![0.42 = \frac{[x]^2}{0.52-x}}](https://tex.z-dn.net/?f=0.42%20%3D%20%5Cfrac%7B%5Bx%5D%5E2%7D%7B0.52-x%7D%7D)
0.42(0.52-x) = x²
0.2184 - 0.42x = x²
x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)
Using the quadratic formula;
; ( where +/- represent ± )
= 
= 
=
OR 
=
OR 
=
OR 
= 0.30225 OR - 0.72225
So, we go by the +ve integer that says:
x = 0.30225
x = [
] = [
] = 0.3023 M
∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).
pH=6.98
Explanation:
This is a very interesting question because it tests your understanding of what it means to have a dynamic equilibrium going on in solution.
As you know, pure water undergoes self-ionization to form hydronium ions, H3O+, and hydroxide anions, OH−.
2H2O(l]⇌H3O+(aq]+OH−(aq]→ very important!
At room temperature, the value of water's ionization constant, KW, is equal to 10−14. This means that you have
KW=[H3O+]⋅[OH−]=10−14
Since the concentrations of hydronium and hydroxide ions are equal for pure water, you will have
[H3O+]=√10−14=10−7M
The pH of pure water will thus be
pH=−log([H3O+])
pH=−log(10−7)=7
Now, let's assume that you're working with a 1.0-L solution of pure water and you add some 10
<span>2.40 - 1.68 =0.72 g of oxigen
moles = 0.72/16 g/mol=0.045
moles x = 1.68/ 55.9=0.03
0.03/0.03 = 1 = x
0.045 / 0.03 = 1.5 = O
to get whole numbers multiply by 2
x2O3
X2O3 +3 CO = 2 X + 3 CO2</span>
Answer:
Check the explanation
Explanation:
When talking about our universe there are 5 d orbitals. The element of first transition series moves away from the universal principles of Hund's rule and Aufbav's principle. So in order to attain stability these elements tend to form half or full filled orbitals.
In our universe the ground state electronic configuration of sixth transition metal, Iron (Fe) : [Ar] 
and the electronic configuration of seventh transition metal, Cobalt (Co) : [Ar] 
=================================
=================================
In universe L there are seven orbitals.
Ground state electronic configuration of sixth and seven transition element.
Sixth transition metal: [Ar] ![3d^{7} 4s^1 or [X] 3d^{7} 4s^1](https://tex.z-dn.net/?f=3d%5E%7B7%7D%204s%5E1%20or%20%5BX%5D%203d%5E%7B7%7D%204s%5E1)
Seventh transition metal: [Ar] ![3d^{7} 4s^{2}or [X] 3d^{7} 4s^{2}](https://tex.z-dn.net/?f=3d%5E%7B7%7D%204s%5E%7B2%7Dor%20%5BX%5D%203d%5E%7B7%7D%204s%5E%7B2%7D)
A. soluble.
The are solubility rules that predict precipitation reaction.