Answer:
a)CH₄, BH₃, and CCl₄
Explanation:
<u>London dispersion forces:-
</u>
The bond for example, in the molecule is F-F, which is non-polar in nature because the two fluorine atoms have same electronegativity values.
The intermolecular force acting in the molecule are induced dipole-dipole forces or London Dispersion forces / van der Waals forces which are the weakest intermolecular force.
Out of the given options, H₂O , NH₃ exhibits hydrogen bonding which is:-
<u>Hydrogen bonding:-
</u>
Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.
Thus option B and C rules out.
<u>Hence, the correct option which represents the molecules which would exhibit only London forces is:- a)CH₄, BH₃, and CCl₄</u>
Answer:
ΔS=0.148 KJ/K
Explanation:
Given that
Q = 100 KJ
T₁=200°C
T₁=200+273 = 437 K
T₂=5°C
T₂=5 + 273 = 278 K
Reservoir 1 is rejecting heat that is why it taken as negative while the reservoir 2 is gaining the heat that is why it is taken as positive.
So the total change in entropy given as
ΔS= - Q/T₁ + Q/T₂
ΔS= - 100/473 + 100/278 KJ/K
ΔS=0.148 KJ/K
Hypothesis because law don’t make sense and theory is for something that already has data behind it. In this case you don’t do it’s not hypothesis
Answer:
A) 2.69 M
B) 0.059
Explanation:
A) We have:
33.8% solute by mass= 33.8 g solute/100 g solution
molarity = mol solute/ 1 L solution
molarity=
x
x
x 
molarity= 2.69 mol solute/L solution = 2.69 M
B) We know that there are 33.8 g of solute in 100 g of solution.
As the total solution is compounded by solute+solvent (in this case, solvent is water), the mass of water is the difference between the mass of the total solution and the mass of solute:
mass of water= 100 g - 33.8 g = 66.2 g
Now, we calculate the number of mol of both solute and water:
mol solute= 33.8 g solute x
= 0.232 mol
mol H20= 66.2 g H₂O x 
Finally, the mol fraction of solute (Xsolute) is calculated as follows:
Xsolute=
Xsolute= 0.059