Explanation:
It is given that,
The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. It means that,
The amount of energy change during the transition is given by :
![\Delta E=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]](https://tex.z-dn.net/?f=%5CDelta%20E%3DR_H%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7Bn_i%5E2%7D%5D)
And
![\dfrac{hc}{\lambda}=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]](https://tex.z-dn.net/?f=%5Cdfrac%7Bhc%7D%7B%5Clambda%7D%3DR_H%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7Bn_i%5E2%7D%5D)
Plugging all the values we get :
![\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{3745\times 10^{-9}}=2.179\times 10^{-18}[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\\dfrac{5.31\times 10^{-20}}{2.179\times 10^{-18}}=[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\0.0243=[\dfrac{1}{n_f^2}-\dfrac{1}{64}]\\\\0.0243+\dfrac{1}{64}=\dfrac{1}{n_f^2}\\\\0.039925=\dfrac{1}{n_f^2}\\\\n_f^2=25\\\\n_f=5](https://tex.z-dn.net/?f=%5Cdfrac%7B6.63%5Ctimes%2010%5E%7B-34%7D%5Ctimes%203%5Ctimes%2010%5E8%7D%7B3745%5Ctimes%2010%5E%7B-9%7D%7D%3D2.179%5Ctimes%2010%5E%7B-18%7D%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7B8%5E2%7D%5D%5C%5C%5C%5C%5Cdfrac%7B5.31%5Ctimes%2010%5E%7B-20%7D%7D%7B2.179%5Ctimes%2010%5E%7B-18%7D%7D%3D%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7B8%5E2%7D%5D%5C%5C%5C%5C0.0243%3D%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7B64%7D%5D%5C%5C%5C%5C0.0243%2B%5Cdfrac%7B1%7D%7B64%7D%3D%5Cdfrac%7B1%7D%7Bn_f%5E2%7D%5C%5C%5C%5C0.039925%3D%5Cdfrac%7B1%7D%7Bn_f%5E2%7D%5C%5C%5C%5Cn_f%5E2%3D25%5C%5C%5C%5Cn_f%3D5)
So, the final level of the electron is 5.
Answer:
B. A precipitate will form since Q > Ksp for calcium oxalate
Explanation:
Ksp of CaC₂O₄ is:
CaC₂O₄(s) ⇄ Ca²⁺ + C₂O₄²⁻
Where Ksp is defined as the product of concentrations of Ca²⁺ and C₂O₄²⁻ in equilibrium:
Ksp = [Ca²⁺][C₂O₄²⁻] = 2.27x10⁻⁹
In the solution, the concentration of calcium ion is 3.5x10⁻⁴M and concentration of oxalate ion is 2.33x10⁻⁴M.
Replacing in Ksp formula:
[3.5x10⁻⁴M][2.33x10⁻⁴M] = 8.155x10⁻⁸. This value is reaction quotient, Q.
If Q is higher than Ksp, the ions will produce the precipitate CaC₂O₄ until [Ca²⁺][C₂O₄²⁻] = Ksp.
Thus, right answer is:
<em>B. A precipitate will form since Q > Ksp for calcium oxalate</em>
<em></em>
Yes, this is balanced. Each part of the substance (like Mg) has the same number on both sides.
Hope this helps you:)
