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:
covalent
Explanation:
The carbon and the nitrogen very often form bonds in nature, carbon-nitrogen bonds, which are covalent types of bonds. In fact, the bonds between the carbon and nitrogen are one of the most abundant in the biochemistry and the organic chemistry. The bonds between these two can be double bonds, as well as triple bonds. The carbon-nitrogen bonds have the tendency to be strongly polarized toward the nitrogen.
I think it’s light
Have fun :)
117 L. You can start by making a table to organize the information you are given. Then, you can use the formula PV/T=PV/T and plug in the numbers you have. You then solve for the missing volume. Remember that the initial pressure, temperature, and volume should be on one side of the equal sign, and the final pressure, volume, and temperature should be on the other side.
