It can allow the molecule (like water) to be polar because it has a negative and positive side to it (oxygen holds the negatives tight causing the hydrogens to be positive).
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:
2.03 moles of Gold
Explanation:
Gold is one of the most precious metal metal used in many applications and mainly as a jewellery. In terms of purity it is categorized in Karats. 24 Karat is considered the purest Gold (i.e. 100 % Gold) while other Karats (14, 18, 22 e.t.c) are alloys with other metals and gyms.
Data Given:
Mass of Gold = 400 g
A.Mass of Gold = 196.97 g.mol⁻¹
Calculate Moles of Gold as,
Moles = Mass ÷ M.Mass
Putting values,
Moles = 400 g ÷ 196.97 g.mol⁻¹
Moles = 2.03 moles of Gold
