Facilitated diffusion is your answer
The correct answer is Cl₂ <CCl₄ < PCl₃ < CaCl₂ < CsCl (most polar)
Cl₂ is nonpolar or least polar as there is no differentiation in electronegativity of the two atoms of chlorine producing amongst them. CCl₄ exhibits four C-Cl bonds. The C-atom is electropositive and Cl is electronegative. Therefore, there are four dipoles in CCl₄. Though, these dipoles cancel each other because of the tetrahedral geometry of the molecule. Therefore, CCl₄ is less polar in comparison to others, however, more polar in comparison to Cl₂.
PCl₃ exhibits three dipoles because of electronegative chlorine atoms, and electropositive phosphorus atom. Though, the configuration of the molecule is trigonal pyramidal, producing it more polar in comparison to CCl₄. The CaCl₂ is ionic compound, therefore, polar in comparison to the covalent bond compounds.
CsCl is ionic compound, however, more polar in comparison to CaCl₂, as there is the higher difference in electronegativity of Cs and Cl compared with that between the Ca and Cl.
Answer : The 
must be administered.
Solution :
As we are given that a vial containing radioactive selenium-75 has an activity of 
.
As, 3.0 mCi radioactive selenium-75 present in 1 ml
So, 2.6 mCi radioactive selenium-75 present in 
Conversion :
Therefore, the must be administered.
Answer:
ΔU = 103.54 KJ
Explanation:
∴ ΔU = Q + W
ideal gas:
∴ PV = nRT
∴ R = 8.314 L.KPa/K.mol
∴ n = 3000 g N2 * ( mol/28.0134g N2) = 107.143 mol N2
∴ m N2 = 3.0 Kg
∴ T1 = 300 K
∴ P1 = 100 KPa
∴ V1 = nRT1/P1 = 2672.36 L = 2.67 m³
⇒ V2 = 0.9*V1 = 2405.12 L = 2.41 m³
∴ P2 = 140 KPa
⇒ T2 = P2.V2/n.R = 377.99 ≅ 378 K
⇒ W = P1V1 - P2V2
⇒ W = ((100KPa)*(2.67m³)) - ((140KPa)*(2.41m³))
⇒ W = - 70.164 KJ
∴ Q = nCpΔT
∴ Cp = (5/2)*R = 20.785 J/mol.K ....ideal gas
⇒ Q = (107.143mol)*(20.785 J/mol.K)*(378 - 300)
⇒ Q = 173703.446 J = 173.703 KJ
⇒ ΔU = 173.703 KJ - 70.164 KJ
⇒ ΔU = 103.54 KJ
