It's lone a little distinction (103 degrees versus 104 degrees in water), and I trust the standard rationalization is that since F is more electronegative than H, the electrons in the O-F bond invest more energy far from the O (and near the F) than the electrons in the O-H bond. That moves the powerful focal point of the unpleasant constrain between the bonding sets far from the O, and thus far from each other. So the shock between the bonding sets is marginally less, while the repugnance between the solitary matches on the O is the same - the outcome is the edge between the bonds is somewhat less.
Somewhere in the Orange to red range
Answer:
+1.03 V
Explanation:
The standard emf of the voltaic cell is the value of the standard potential of it, which is calculated by the standard reduction potential (E°).
The standard reduction potential is the potential needed for the reduction reaction happen, and it's determined by the reaction with the hydrogen cell (which has E° = 0.0V). The half-reactions of reduction of Ni⁺² and Ag⁺, are:
Ni⁺²(aq) + 2e⁻ → Ni(s) E° = -0.23 V
Ag⁺(aq) + e⁻ → Ag(s) E° = +0.80 V
The value is calculated by a spontaneous reaction, in which the cell with the greater E° is reduced (gain electrons), and the other is oxidized (loses electrons). So, Ag⁺ reduces.
emf = E°reduces - E°oxides
emf = 0.80 - (-0.23)
emf = +1.03 V
<span>The answer is hypertonic. In osmosis, water
molecules move from a hypotonic solution to the hypertonic solution, through a
semipermeable membrane. This occurs until
both solutions become isotonic relative to each other. In osmosis, only
the movement of water molecules occurs since the ions are large enough to pass
through the pores of the semipermeable membrane,
in this case, the cell membrane. Due to
loss of water in the process of osmosis, the cells in the fingers of the swimmers
shrunk hence looked shriveled.</span>
