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.
The Mr is the mass numbers of each element added up so…. Fe = 56, O=16, H=1 … now add these up with the number of each element -> there’s 1 Fe, and 3 Os and 3 Hs as they are in brackets with a 3 outside-> (56+16+16+16+1+1+1=107) … your answer is 107
Answer:
see below
Explanation:
The rate constant is missing in question, but use C(final) = C(initial)e^-kt = 0.200M(e^-k·10). Fill in k and compute => remaining concentration of reactant
Answer:
14) The edge dislocation is more plastic than the screw dislocation
15) So as to form kinks that are fast moving
Explanation:
14) Edge and screw dislocations are the two main types of mobile dislocations
The three dimensional core of the screw dislocation prevents the slipping of the layers (one over the other) in a BCC metal such that kinks are required to be formed first by thermal activation (heating) in order. The kinks are edge dislocation that move such that the screw dislocation moves forward
Hence, the edge dislocation is more plastic than the screw dislocation
15) The three dimensional structure of a screw dislocation acts like a wedge which resists the slipping of the layers in the BCC structure such that the screw dislocation needs to be highly thermally activated forming kinks before the surrounding layers can move.
Answer:
<em>293.99 g </em>
OR
<em>0.293 Kg</em>
Explanation:
Given data:
Lattice energy of Potassium nitrate (KNO3) = -163.8 kcal/mol
Heat of hydration of KNO3 = -155.5 kcal/mol
Heat to absorb by KNO3 = 101kJ
To find:
Mass of KNO3 to dissolve in water = ?
Solution:
Heat of solution = Hydration energy - Lattice energy
= -155.5 -(-163.8)
= 8.3 kcal/mol
We already know,
1 kcal/mol = 4.184 kJ/mole
Therefore,
= 4.184 kJ/mol x 8.3 kcal/mol
= 34.73 kJ/mol
Now, 34.73 kJ of heat is absorbed when 1 mole of KNO3 is dissolved in water.
For 101 kJ of heat would be
= 101/34.73
= 2.908 moles of KNO3
Molar mass of KNO3 = 101.1 g/mole
Mass of KNO3 = Molar mass x moles
= 101.1 g/mole x 2.908
= 293.99 g
= 0.293 kg
<em><u>293.99 g potassium nitrate has to dissolve in water to absorb 101 kJ of heat. </u></em>
