The answer to your question is a 12
The answer is The loudness of sound is related to its amplitude, this is off edmentum exactly so I advise changing up the wording. You can say something about the pitch or you can word it like, The sound of the wave is related to how loud the sound can be. Hope this helped
Answer:
3.07 × 10⁻⁴
Explanation:
Step 1: Calculate the concentration of H⁺
We will use the definition of pH.
![pH = -log [H^{+} ]\\\[ [H^{+} ] = antilog -pH = antilog -2.37 = 4.27 \times 10^{-3} M](https://tex.z-dn.net/?f=pH%20%3D%20-log%20%5BH%5E%7B%2B%7D%20%5D%5C%5C%5C%5B%20%5BH%5E%7B%2B%7D%20%5D%20%3D%20antilog%20-pH%20%3D%20antilog%20-2.37%20%3D%204.27%20%5Ctimes%2010%5E%7B-3%7D%20M)
Step 2: Calculate the concentration of HY
5.22 × 10⁻³ mol of HY are dissolved in 0.088 L. The concentration of the acid (Ca) is:
Step 3: Calculate the acid dissociation constant (Ka)
We will use the following expression.
![Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(4.27 \times 10^{-3} )^{2} }{0.0593} = 3.07 \times 10^{-4}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%20%7D%7BCa%7D%20%3D%20%5Cfrac%7B%284.27%20%5Ctimes%2010%5E%7B-3%7D%20%29%5E%7B2%7D%20%7D%7B0.0593%7D%20%3D%203.07%20%5Ctimes%2010%5E%7B-4%7D)
Answer:
None of the above.
Explanation:
The limiting factor of this is the chlorine. You only have 2 moles of chlorine.
2Na + Cl_2 ==> 2*NaCl
The equation tells you that for every mol of Cl2 that you have, you require 2 moles of Na and you get 2 moles of NaCl
So what that means is that if you have 2 mols of Cl2, you need just 4 moles of Na, and you will get 4 mols of NaCl
Since 4 is not one of your choices, the answer is none of the above
