Answer is: ammonia experience only dispersion intermolecular forces with BF₃ (boron trifluoride) because BF₃ is only nonpolar molecule (vectors of dipole moments cansel each other, dipole moment is zero).
The London dispersion force (intermolecular force) <span>is a temporary attractive </span>force between molecules.
The point at which one drop of base turns the acid indicator into a pink color that lasts for thirty seconds in doing titration is called the end point or the equivalence point.
End point or the equivalence point is the one responsible for the pink color that lasts for thirty seconds.
Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol
Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol
Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
= (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
= 176.15 kJ - 84.78 kJ = 91.38 kJ
Answer:
.
Explanation:
Electrons are conserved in a chemical equation.
The superscript of 
indicates that each of these ions carries a charge of 
. That corresponds to the shortage of one electron for each 
ion.
Similarly, the superscript 
on each 
ion indicates a shortage of three electrons per such ion.
Assume that the coefficient of (among the reactants) is 
, and that the coefficient of (among the reactants) is 
.
.
There would thus be silver (
) atoms and aluminum (
) atoms on either side of the equation. Hence, the coefficient for 
and 
would be 
and 
, respectively.
.
The ions on the left-hand side of the equation would correspond to the shortage of electrons. On the other hand, the 
ions on the right-hand side of this equation would correspond to the shortage of 
electrons.
Just like atoms, electrons are also conserved in a chemical reaction. Therefore, if the left-hand side has a shortage of electrons, the right-hand side should also be electrons short of being neutral. On the other hand, it is already shown that the right-hand side would have a shortage of electrons. These two expressions should have the same value. Therefore, 
.
The smallest integer and that could satisfy this relation are 
and 
. The equation becomes:
.
