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:
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Explanation:
Answer:
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You need to find the abundance. Then, multiply the abundance by 100, and add that to the mass for each isotope. Basically, for each isotope, take the percentage abundance and add it to the mass. Multiply each calculation of these together to get your average atomic mass,
Option 2: 12.0 L of 
at STP.
The standard pressure and temperature values are 1 atm and 273.15 K.
Using the ideal gas equation, number of moles of gas can be calculated which is as follows:
PV=nRT...... (1)
Here, P is pressure, V is volume, n is number of moles, R is gas constant and T is temperature.
Also, in 1 mole of any gas there are 
molecules of the gas. This is known as Avogadro's number and denoted by symbol 
Thus,
Equation (1) can be rewritten as follows:
On rearranging,
Here, all the other terms are constant except volume, thus, gas with volume equal to the volume of 
will have same number of molecules.
Volume of gas and 
gas is same thus, will have same total number of molecules as gas.
