Answer:
Born-Haber Process - "Born-Haber process applies Hess's law to calculate the lattice enthalpy(<em>this is a thermodynamic term which equals to the total heat content of a system</em>), by comparing the standard enthalpy change of formation of ionic compound (from the elements) to the enthalpy required to make gaseous ions from the elements".
Explanation:
Now the calculations done for SRI2 are as follows:
- Enthalpy of Sublimation of Sr(s)= 164 kJ per mol,
- 1st ionization energy of Sr(g)=549 kJ per mol,
- 2nd ionization energy of Sr(g)=1064 kJ per mol,
- Enthalpy dissociation energy of I2(s)= 62.4 kJ per mol,
- Bond dissociation energy of I2(g)= 152.55 kJ per mol,
- 1st electron affinity of I(g)= -295.15 kJ per mol,
- Lattice energy of SrI2(s)=-1959.75 kJ per mol.
Note - the terms mentioned in the brackets are as, S- solid form, g- Gaseous form.
Answer:
2360J
Explanation:
Given parameters:
Mass of aluminum = 25g
Melting point = 658°C
Latent heat of melting = 94.4j/g
Unknown:
Amount of heat require for melting = ?
Solution:
The amount of heat required for this melting will be the product of the mass of the aluminum and the latent heat of melting;
H = m L
m is the mass
L is the latent heat of melting
Insert the parameters and solve;
H = 25 x 94.4 = 2360J
Answer:
my guy we need to read lesson cuz int know what any of that mean
Explanation:
Answer:
10.3 g Al
Explanation:
To find the excess mass of Al, you need to (1) convert grams I₂ to moles I₂ (via molar mass), then (2) convert moles I₂ to moles AlI₃ (via mole-to-mole ratio from equation coefficients), then (3) convert moles AlI₃ to grams AlI₃ (via molar mass). Now that you have the actual amount of AlI₃ produced, you need to (4) convert grams AlI₃ to moles AlI₃ (via molar mass), then (5) convert moles AlI₃ to moles Al (via mole-to-mole ratio from equation coefficients), and then (6) convert moles Al to grams Al (via molar mass). Now that you know the amount of Al actually needed to produce the product, you need to (7) find the excess mass of Al.
Molar Mass (Al): 26.982 g/mol
Molar Mass (I₂): 2(126.90 g/mol)
Molar Mass (I₂): 253.8 g/mol
Molar Mass (AlI₃): 26.982 g/mol + 3(126.90 g/mol)
Molar Mass (AlI₃): 407.682 g/mol
2 Al(s) + 3 I₂(g) -------> 2 AlI₃(g)
113 g I₂ 1 mole 2 moles AlI₃ 407.682 g
------------- x ---------------- x ---------------------- x ------------------- = 121 g AlI₃
253.8 g 3 moles I₂ 1 mole
121 g AlI₃ 1 mole 2 moles Al 26.982 g
--------------- x ------------------ x --------------------- x ----------------- = 8.01 g Al
407.682 g 2 moles AlI₃ 1 mole
Starting Amount - Mass Needed = Excess
18.3 g Al - 8.01 g Al = 10.3 g Al