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
The correct answer is option 1. Butane and 2-butene have the same total number of carbon atoms. They both have four carbon atoms. They differ in there structure since the latter has double bonds on it. As a result of the different structure, they also have different properties.
Hydrogen and Oxygen by themselves have 1 and 6 valence electrons, respectively. 8 valence electrons is stable, so the atoms form bonds with each other to achieve 8 valence e-.
1 H atom + 1 H atom + 1 O atom = 8 valence e-
<span>pm stands for picometer and picometers are units which can be used to measure really tiny distances. One picometer is equal to 10^{-12} meters. We know that one centimeter is equal to 10^{-2} m so there are 10^2 cm per meter.
We can change the distance d = 115 pm to units of centimeters.
d = (115 pm) x (10^{-12}m / pm) x (10^2 cm / m)
d = 115 x 10^{-10} cm = 1.15 x 10^{-8} cm
The distance in centimeters is 1.15 x 10^{-8} cm</span>
