Answer:
The correct answer is: Ka= 5.0 x 10⁻⁶
Explanation:
The ionization of a weak monoprotic acid HA is given by the following equilibrium: HA ⇄ H⁺ + A⁻. At the beginning (t= 0) we have 0.200 M of HA. Then, a certain amount (x) is dissociated into H⁺ and A⁻, as is detailed in the following table:
HA ⇄ H⁺ + A⁻
t= 0 0.200 M 0 0
t -x x x
t= eq 0.200M -x x x
At equilibrium, we have the following ionization constant expression (Ka):
Ka= ![\frac{ [H^{+}] [A^{-} ]}{ [HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%5BH%5E%7B%2B%7D%5D%20%20%5BA%5E%7B-%7D%20%5D%7D%7B%20%5BHA%5D%7D)
Ka= 
Ka= 
From the definition of pH, we know that:
pH= - log [H⁺]
In this case, [H⁺]= x, so:
pH= -log x
3.0= -log x
⇒x = 10⁻³
We introduce the value of x (10⁻³) in the previous expression and then we can calculate the ionization constant Ka as follows:
Ka= 
= 
= 5.025 x 10⁻⁶= 5.0 x 10⁻⁶
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Answer:
0.1 mole of CH₄
Explanation:
From the question given above, the following data were obtained:
Volume of CH₄ = 2.24 L
Number of mole of CH₄ =?
The number of mole of CH₄ can be obtained as follow:
Recall:
1 mole of a gas occupy 22.4 L at stp. This implies that 1 mole of CH₄ occupies 22.4 L at stp.
22.4 L = 1 mole of CH₄
Therefore,
2.24 L = 2.24 × 1 mole of CH₄ / 22.4
2.24 L = 0.1 mole of CH₄.
Answer:
25.8
Explanation:
Let's write the reaction between magnesium-phosphide and potassium:
Mg3P2 + K = Mg + K3P
And now let's balance this equation:
Mg3P2+6K=3Mg+2K3P
We see that the ratio of magnesium-phosphide and potassium is 1:6, which means that for every mole of magnesium-phosphide there need to be 6 moles of potassium.
Since we have 4.3 moles of Mg3P2, there need to be 6 • 4.3 = 25.8 moles of potassium.
