The Lyman series can be expressed in the formula <span><span>1/λ</span>=<span>RH</span><span>(1−<span>1/<span>n2</span></span>) where </span><span><span>RH</span>=1.0968×<span>107</span><span>m<span>−1</span></span>=<span><span>13.6eV</span><span>hc
</span></span></span></span>Where n is a natural number greater than or equal to 2 (i.e. n = 2,3,4,...). Therefore, the lines seen in the image above are the wavelengths corresponding to n=2 on the right, to n=∞on the left (there are infinitely many spectral lines, but they become very dense as they approach to n=∞<span> (Lyman limit), so only some of the first lines and the last one appear).
The wavelengths (nm) in the Lyman series are all ultraviolet
:2 3 4 5 6 7 8 9 10 11
Wavelength (nm) 121.6 102.6 97.3 95 93.8 93.1 92.6 92.3 92.1 91.9 91.18 (Lyman limit)
In your case for the n=5 line you have to replace "n" in the above formula for 5 and you should get a value of 95 x 10^-9 m for the wavelength. then you have to use the other equation that convert wavelength to frequency. </span>
<u>Answer:</u> The amount of water required to prepare given amount of salt is 398.4 mL
<u>Explanation:</u>
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

We are given:
Molarity of solution = 0.16 M
Given mass of manganese (II) nitrate tetrahydrate = 16 g
Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol
Putting values in above equation, we get:

Volume of water = Volume of solution = 398.4 mL
Hence, the amount of water required to prepare given amount of salt is 398.4 mL
The experimental mole ratio of silver chloride to barium chloride is calculated as below
fin the mole of each compound
mole= mass/molar mass
moles of AgCl = 14.5g/142.5 g/mol = 0.102 moles of AgCl
moles of BaCl2 = 10.2 g/208 g/mol = 0.049 moles of BaCl2
find the mole ratio by dividing each mole with the smallest mole(0.049)
AgCl= 0.102/0.049 =2
BaCl2 = 0.049/0.049 =1
therefore the mole ratio AgCl to BaCl2 is 2 :1