A.) Low frequency and Long Wavelength
Answer: The hydroxide concentration of this sample is 
Explanation:
When an expression is formed by taking the product of concentration of ions raised to the power of their stoichiometric coefficients in the solution of a salt is known as ionic product.
The ionic product for water is written as:
![K_w=[H^+]\times [OH^-]](https://tex.z-dn.net/?f=K_w%3D%5BH%5E%2B%5D%5Ctimes%20%5BOH%5E-%5D)
![7.7\times 10^{-14}=[H^+]\times [OH^-]](https://tex.z-dn.net/?f=7.7%5Ctimes%2010%5E%7B-14%7D%3D%5BH%5E%2B%5D%5Ctimes%20%5BOH%5E-%5D)
As ![[H^+]=[OH^-]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BOH%5E-%5D)
![2[OH^-]=7.7\times 10^{-14}](https://tex.z-dn.net/?f=2%5BOH%5E-%5D%3D7.7%5Ctimes%2010%5E%7B-14%7D)
![[OH^-]=3.85\times 10^{-7}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D3.85%5Ctimes%2010%5E%7B-7%7D)
Thus hydroxide concentration of this sample is 
First let us compute for the number of moles of butane
(molar mass = 58.12 g/mol)
number of moles = 145 g / (58.12 g/mol) = 2.49 mol
<span>We use the ideal
gas equation to calculate the volume:</span>
<span> V = n R T / P</span>
V = 2.49 mol * 62.36367 L torr / mol K * 308.15 K / 745
torr
<span>V = 64.35 L</span>