Q: What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0 °C? (Lf = 3.34 x 105 J/kg)
Answer:
-3670.33 J/K
Explanation:
Entropy: This can be defined as the degree of randomness or disorderliness of a substance. The S.I unit of Entropy is J/K.
Mathematically, change of Entropy can be expressed as,
ΔS = ΔH/T ....................................... Equation 1
Where ΔS = Change of entropy, ΔH = heat change, T = temperature.
ΔH = -(Lf×m).................................... Equation 2
Note: ΔH is negative because heat is lost.
Where Lf = latent heat of ice = 3.34×10⁵ J/kg, m = 3.0 kg, m = mass of water = 3.0 kg
Substitute into equation
ΔH = -(3.34×10⁵×3.0)
ΔH = - 1002000 J.
But T = 0 °C = (0+273) K = 273 K.
Substitute into equation 1
ΔS = -1002000/273
ΔS = -3670.33 J/K
Note: The negative value of ΔS shows that the entropy of water decreases when it is changed to ice at 0 °C
Answer:
In He2 molecule,
Atomic orbitals available for making Molecular Orbitals are 1s from each Helium. And total number of electrons available are 4.
Molecular Orbitals thus formed are:€1s2€*1s2
It means 2 electrons are in bonding molecular orbitals and 2 are in antibonding molecular orbitals .
Bond Order =Electrons in bonding molecular orbitals - electrons in antibonding molecular orbitals /2
Bond Order =Nb-Na/2
Bond Order =2-2/2=0
Since the bond order is zero so that He2 molecule does not exist.
Explanation:
Answer:
They gave you the equation; Cp=,
just plug everything in! You’ve seen this; I have long ago, but we had different units. Sorry, but it’s right there! Go get it!
Explanation:
