For a comparison of the nucleus 5626fe, the density of the nucleus 112 48cd is mathematically given as the same.
n(Cd) / n(Fe)=1
<h3>What is the density of the nucleus 112 48cd?</h3>
Generally, the equation for the density is mathematically given as
d=\frac{A}{4/3}\piR^3
Therefore
n(Cd) / n(Fe) = [A (Cd) / (A Fe) ] * [ R (Fe) / R (Cd)]^3
n(Cd) / n(Fe)= (112 / 56 ) * (1/1.26)3
n(Cd) / n(Fe)=1
In conclusion, The ratio of n(Cd) = n(Fe) is 1, hence same
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brainly.com/question/14010194
Answer:
Kc = 1.54e - 31 / 2.61e - 24
Explanation:
1 )
; Kc = 1.54e - 31
2)
; Kc = 2.16e - 24
upon reversing ( 2 ) equation
Kc = 1/2.16e - 24
now adding 1 and reversed equation (2)


we get ,
Kc = 1.54e-31 × 1/2.61e - 24
equilibrium constant of equation (3) is -
Kc = 1.54e - 31 / 2.61e - 24
Answer:
NH3>H2O>Cl-
Explanation:
The given wavelengths of maximum absorption for each complex can be used to estimate the magnitude of field splitting of the respective ligands as shown in the image attached. The field splitting is reported in the unit kilojoule per mole (KJmol-1).
It can be seen from the calculation in the image attached that ammonia shows the highest crystal field splitting followed by water and lastly the chloride anion. This corresponds to the respective positions of these species in the spectrochemical series. Water and the chloride ion are weak field ligands.
<span>Selenium (se)
........
.....</span>
Answer:
I believe it's 4892,08808J
Explanation:
∆T=158,8g•4,18÷g°C•7,37°C
=4'892,08808J