Can someone please explain what to do it’s late at night and I need sleep but I forgot I had this homework please help me with a
few questions I’m desperate :(
1 answer:
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Answer: 0.0023
Explanation:
I just answered this very same question. You can read my answer here: brainly.com/question/12077289.
1) First, you must write the balanced equilibrium equation, which will let you to determine the mole ratios of the different species in solution:
HF ⇆ H⁺ + F⁻
Therefore, the mole ratio is: 1 HF: 1 H⁺ : 1 F⁻
2) Second, write the equilibrium constant of the acid, Ka, which will permit you to state the relationship of the concentrations in equilibrium:
Ka = [H⁺] [F⁻] / [HF]
[HF] is the searched concentration of the hydrofluoric acid
From the mole ratio of the balanced chemical equation [H⁺] = [F⁻]
Hence:
Ka = [H⁺] [H⁺] / [HF] = [H⁺]² / [HF]
3) From the pH value you can calculate [H⁺], using the definition:
- pH= - log [H⁺]
- Substitute: 2.90 = - log [H⁺]
Clear [H⁺] using logarithm properties:
4) Now you can substitute [H⁺] and the value of Ka in the equation for Ka:
Ka = [H⁺]² / [HF] ⇒ [HF] = [H⁺]² / Ka = (0.00126)² / (6.8×10⁻⁴) = 0.00233.
Since the value of Ka has two significant figures, you must report the answer with two significan figures, i.e. 0.0023
The percentage of excess air used during combustion process of ethane will be 37 %.
Burning, also known as combustion, would be a high-temperature highly exothermic chemical process that occurs when an oxidant, typically atmospheric oxygen, interacts with a fuel to generate oxidized, frequently gaseous products in a mixture known as smoke.
Calculation of percentage of air .
Mair=Mair/Rin
+
÷
+
+
33 . 3.25(1-x) + 28 × 13.16(1-x) ÷ 33 × 3.25(1-x) + 28 × 13.16(1-x). + 30.1
= 176/176+8
X= 0.37
0.37 × 100
X= 37%
Therefore, the percentage of excess air used during combustion process of ethane will be 37 %.
To know more about combustion process
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