Answer: It is because tyrosine kinases and BTK have similar solubilities
Explanation:
In column chromatography, components of a mixture are seperated based on their relative solubilities in two non-mixing phases.
In essence, tyrosine kinases and BTK are present in the eluate due to their similar solubility rates that arise from the similar chemical structure both possess (otherwise it would be impossible for the inhibitor meant for Tyrosine kinase to bind and also inhibits BTK)
Thus, the similar solubilities of both groups is the reason they could elute out of the column without being adsorped.
1) Since you have not provided the equations to select the right one, I am going to explain you the relevant facts that are used to solve this question.
2) The transuranium elements are the chemiical elements with atomic number greater than that of the uranium.
The atomic number of uranium is 92. So, the transuranium elements are the elements with atomic number 93 or greater.
This are some of the transuranium elements:
Neptunio - 93
Plutonium - 94
Americium - 95
Curium - 96
Berkelium - 97
Californium - 98
Einstenium - 99
And so all the known elements (the last one is the 118).
3) In a nuclear reaction the total mass number ( shown as superscript to the left of the symbol) and total atomic number (shown as subscript to the left of the symbol) are conserved.
4) Beta decay is the release of a beta particle, which is an electron (considered massles and with charge - 1). So, the beta decay is represented with the symbol:
0
β, which means 0 mass and charge - 1.
-1
5) This is, then, an example of a β decay equation for one transuranium element:
239 239 0
Np → Pu + β
93 94 -1
As you see 239 = 239 + 0 and 93 = 94 - 1, showing that the total mass number ( shown as superscript to the left of the symbol) and the total atomic number (shown as subscript to the left of the symbol) are conserved.
the molecular formula is calculated as follows
find the mole of each molecule
that is for carbon = 26.7/12= 2.23 moles
hydrogen= 2.24/1=2.24 moles
oxygen= 71.1/16=4.44 moles
find the mole ratio of each element
that is divide all moles by the smallest mole( 2.23)
for carbon=2.23/2.23=1,
hydrogen = 2.24/2,23=1
oxygen= 4.44/2.23= 2
the empirical formula = CHO2
therefore the molecular formula=(CHO2)n=270.1
{ ( 12x 1) + 1x1) +(16x2)}n=270.1
45n=270.1
n=6
molecular formula is therefore=(CHO2)6=C6H6O12
Answer: The standard enthalpy of formation of this isomer of 
is -210.9 kJ
Explanation:
The given balanced chemical reaction is,
First we have to calculate the enthalpy of formation of 
.
![\Delta H^o=[n_{CO_2}\times \Delta H_f^0_{(CO_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{O_2}\times \Delta H_f^0_{(O_2)+n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28CO_2%29%7D%2Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%5D-%5Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%2Bn_%7BC_8H_%7B18%7D%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%7D%5D)
where,
We are given:
Putting values in above equation, we get:
![-511.3kJ/mol=[(8\times -393.5)+(9\times -241.8)]-[(\frac{25}{2}\times 0)+(1\times \Delta H_f^0_{(C_8H_{18})}](https://tex.z-dn.net/?f=-511.3kJ%2Fmol%3D%5B%288%5Ctimes%20-393.5%29%2B%289%5Ctimes%20-241.8%29%5D-%5B%28%5Cfrac%7B25%7D%7B2%7D%5Ctimes%200%29%2B%281%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%7D)
Hawaii, Iceland, and Yellowstone are examples of places that were formed by hot spot volcanoes.
Hope this helps!
