Explanation:
Given the mass of HCl is ---- 0.50 g
The volume of solution is --- 4.0 L
To determine the pH of the resulting solution, follow the below-shown procedure:
1. Calculate the number of moles of HCl given by using the formula:
2. Calculate the molarity of HCl.
3. Calculate pH of the solution using the formula:
![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
Since HCl is a strong acid, it undergoes complete ionization when dissolved in water.
Thus, ![[HCl]=[H^+]](https://tex.z-dn.net/?f=%5BHCl%5D%3D%5BH%5E%2B%5D)
Calculation:
1. Number of moles of HCl given:
2. Concentration of HCl:
3. pH of the solution:
![pH=-log[H^+]\\=-log(0.003425)\\=2.47](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D%5C%5C%3D-log%280.003425%29%5C%5C%3D2.47)
Hence, pH of the given solution is 2.47.
The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>
the diagram of the emission spectrum has been added.
<em>From the given</em><em> chart;</em>
The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m
The frequency of this emission is calculated as follows;
c = fλ
where;
- <em>c is the speed of light = 3 x 10⁸ m/s</em>
- <em>f is the frequency of the wave</em>
- <em>λ is the wavelength</em>
The energy of the emitted photon corresponding to the orange line is calculated as follows;
E = hf
where;
- <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>
<em />
E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)
E = 3.26 x 10⁻¹⁹ J.
Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
Learn more here:brainly.com/question/15962928
The full question is shown in the image attached
Answer:
See explanation
Explanation:
In naming an alkane, the first thing we do is to obtain the parent chain by counting the number of carbon atoms in the chain.
When we obtain that, then we identify the substituents and number them in such a way that they have the lowest numbers. The compounds shown have the following names according to the order in which the structures appear in the image attached;
1. 2-methyl propane
2. 2,4-dimethyl heptane
3. 2,2,3,3-tetramethyl butane
4. 5-ethyl-2,4-dimethyl octane
