Answer : The enthalpy change for converting 1 mole of ice at 
to water at 
is, 7.712 KJ
Solution :
Process involved in the calculation of enthalpy change :
Now we have to calculate the enthalpy change.
![\Delta H=[m\times c_{ice}\times (T_2-T_1)]+\Delta H_{fusion}+[m\times c_{water}\times (T_3-T_2)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bice%7D%5Ctimes%20%28T_2-T_1%29%5D%2B%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bwater%7D%5Ctimes%20%28T_3-T_2%29%5D)
where,
= enthalpy change
m = mass of water = 
= specific heat of ice = 2.09 J/gk
= specific heat of water = 4.18 J/gk
= enthalpy change for fusion = 6.01 KJ/mole = 0.00601 J/mole
conversion : 
= initial temperature of ice =
= final temperature of ice = 
= initial temperature of water =
= final temperature of water = 
Now put all the given values in the above expression, we get
![\Delta H=[18g\times 2.09J/gK\times (273-248)k]+0.00601J+[18g\times 4.18J/gK\times (363-273)k]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B18g%5Ctimes%202.09J%2FgK%5Ctimes%20%28273-248%29k%5D%2B0.00601J%2B%5B18g%5Ctimes%204.18J%2FgK%5Ctimes%20%28363-273%29k%5D)
(1 KJ = 1000 J)
Therefore, the enthalpy change for converting 1 mole of ice at to water at is, 7.712 KJ
Answer:
Exercise 1
a) 140 m
b) 100 m
c) 180 m
d) 140 m
Exercise 2
a) 20 yards
b) 30 yards
c) 20 yards
d) 55 yards
Exercise 3
a) 11 Kilometers
b) 7 Kilometers
Explanation:
Exercise 1
Distance from B to C = 140 m
Distance from to D = 100 m
Total distance = 100+40+40 = 180 m
Total displacement i.e. distance between A and D is 140 m
Exercise 2
a) Distance from B to C = 35 -15 = 20 yards
b) Distance from C to D = 5 + 35 = 30 yards
c) Distance from B to D = 5 + 15 = 20 yards
d) Displacement = 5 + 50 = 55 yards
Exercise 3
a) The total distance travelled = 5 + 2 + 4 = 11 Kilometers
b) Displacement = 5-2 + 4 = 7 Kilometers
The salt makes it rise and
Answer:
CN^- is a strong field ligand
Explanation:
The complex, hexacyanoferrate II is an Fe^2+ specie. Fe^2+ is a d^6 specie. It may exist as high spin (paramagnetic) or low spin (diamagnetic) depending on the ligand. The energy of the d-orbitals become nondegenerate upon approach of a ligand. The extent of separation of the two orbitals and the energy between them is defined as the magnitude of crystal field splitting (∆o).
Ligands that cause a large crystal field splitting such as CN^- are called strong field ligands. They lead to the formation of diamagnetic species. Strong field ligands occur towards the end of the spectrochemical series of ligands.
Hence the complex, Fe(CN)6 4− is diamagnetic because the cyanide ion is a strong field ligand that causes the six d-electrons present to pair up in a low spin arrangement.
