Answer:
- 5.15×10²⁴ molecules of sulfur dioxide
- 3.63×10²³ molecules of carbon monoxide
- 6.02×10²³ molecules of ammonia
Explanation:
We begin from the relation that 1 mol of molecules contains NA of molecules
NA = 6.02×10²³
Now, we make rules of three:
1 mol has 6.02×10²³ molecules, therefore:
8.55 moles of SO₂ must have (8.55 . NA) / 1 = 5.15×10²⁴ molecules of dioxide
0.603 moles of CO must have (0.603 . NA) / 1 = 3.63×10²³ molecules of monoxide
Avogadro's Number of molecules of NH₃ are 6.02×10²³ molecules of ammonia
<span>The atomic weight of 13C should be pretty close to 13.0. (If you have the exact mass, use it in the problem.) So,
9.00 g / 13.0 g/mol = 0.692 moles
Therefore, the answer should be 0.692 moles are in 9.00 g of 13C.</span>
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:
I hope this is it. I'm not really sure.
