Strength, the capacity of the invi elements
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>
Answer:
4) 0.26 atm
Explanation:
In the process:
Benzene(l) → Benzene(g)
ΔG° for this process is:
ΔG° = -RT ln Q
<em>Where Q = P(Benzene(g)) / P°benzene(l) P° = 1atm</em>
ΔG° = 3700J/mol = -8.314J/molK * (60°C + 273.15) ln P(benzene) / 1atm
1.336 = ln P(benzene) / 1atm
0.26atm = P(benzene)
Right answer is:
<h3>4) 0.26 atm
</h3><h3 />
a. t=0.553 s
b. vox(horizontal speed) = 3.62 m/s
<h3>Further explanation</h3>
Given
h = 1.5 m
x = 2 m
Required
a. time
b. vo=initial speed
Solution
Free fall motion
a. h = 1/2 gt²(vertical motion=h=voyt+1/2gt²⇒voy = 0)
t = √2h/g
t = √2.1.5/9.8
t=0.553 s
b. x=vox.t(horizontal motion)
vox=x/t
vox=2/0.553
vox=3.62 m/s
D as it only has a negative effect when being used and it is small at that anyway.
