Answer:
Compound 1 is molecular
Compound 2 is ionic
Compound 3 can't really be decided
Explanation:
A molecular substance does not conduct electricity, has very low melting and boiling points and is held together by very weak intermolecular forces.
An ionic substance conducts electricity in solution or in molten state but never in the pure solid state, has a high melting and boiling point and has a dull appearance most times.
Compounds 1 shows the properties of molecular substances hence it are designated as such.
On the other hand, compound 2 shows the properties of an ionic substance and is also designated as such.
We can't really decide on compound 3 because it shows some properties of ionic substances and some properties of molecular substances.
Answer:
The spin of the complex is 5.92 B.M
Explanation:
Please see the attachments below
The mole fraction of KCl in the solution is 0.1051
calculation
mole fraction of KCl in solution = moles of KCl / total number of moles(moles of KCl +moles of H2O)
moles=mass/molar mass
mass of KCl=32.7g
molar mass of KCl= 39 +35.5
moles of KCl is therefore= 32.7g/74.5 g/mol=0.439 moles
find the moles of H2O= mass of H2O/molar mass
mass of H2O=100-32.7=67.3g
molar mass of H2O=( 1 x2) +16=18 g/mol
moles = 67.3/18 =3.739 moles
total moles=3.739+0.439=4.178 moles
mole fraction is therefore=0.439/4.178=0.1051
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
Cheeze is the most one i hope it helps
