Answer:
a) pH = 9.82 b) pH = 1.65
a) pOH = 7.8 b) pOH = 4.45
Explanation:
pOH + pH = 14 for all of these solutions.
The value of pka for
in an aqueous solution is 9.2.
<h3>What is Kb?</h3>
Kb denotes the base dissociation constant.
pKa + pKb =14 at 25 degree celcius.
pKa + 4.8 =14
pKa = 9.2
pKb is the negative base-10 logarithm of the base dissociation constant (Kb) of a solution.
It is used to determine the strength of a base or alkaline solution.
The value of pka for
in an aqueous solution is 9.2.
Learn more about pka here:
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<h3>
Answer:</h3>
![\displaystyle K_{eq} = \frac{[H_2]^2[O_2]}{[H_2O]^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BH_2%5D%5E2%5BO_2%5D%7D%7B%5BH_2O%5D%5E2%7D)
<h3>
General Formulas and Concepts:</h3>
<u>Chemistry</u>
<u>Aqueous Solutions</u>
<u>Kinetics</u>
<u>Equilibrium</u>
- Equilibrium Reactions RxN ⇄
- Equilibrium Constants
- Equilibrium Expression:
![\displaystyle K = \frac{[Producst]^{coefficients}}{[Reactants]^{coefficients}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K%20%3D%20%5Cfrac%7B%5BProducst%5D%5E%7Bcoefficients%7D%7D%7B%5BReactants%5D%5E%7Bcoefficients%7D%7D)
<h3>
Explanation:</h3>
*Note: Only gases and liquids affect equilibrium
<u>Step 1: Define</u>
[RxN - Balanced] 2H₂O (g) ⇄ 2H₂ (g) + O₂ (g)
<u>Step 2: Identify</u>
[Reactant] H₂O
[Product] H₂
[Product] O₂
<u>Step 3: Write K Expression</u>
- Substitute:
![\displaystyle K_{eq} = \frac{[H_2]^2[O_2]}{[H_2O]^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BH_2%5D%5E2%5BO_2%5D%7D%7B%5BH_2O%5D%5E2%7D)
Answer:
27 mL
Explanation:
Please see the step-by-step solution in the picture attached below.
Hope this answer can help you. Have a nice day!
Answer:
0.185 moles of AlN
Explanation:
The equation is:
2Al + N₂ → 2AlN
This reaction indicates that 2 mol of aluminum can react to 1 mol of nitrogen to produce 2 moles of aluminum nitride.
We convert the aluminum's mass to moles
5 g/ 26.98 g/mol = 0.185 moles.
2 moles of Al are needed to produce 2 moles of AlN
so, ratio is 1:1
In conclussion we say:
0.185 moles of Al, must produce 0.185 moles of AlN