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:
what do you need help with?
Explanation:
The amount of kinetic energu has depens upon an object's speed and its mass.
So its I and IV.
The correct answer is B.
Hope it helps!
#MissionExam001
Because there are so many different values of numbers, it would be impractical to use 1Ω, 2Ω, 3Ω... etc... Using colored bands helps make reading it a little easier to the trained eye. There are hundreds of thousands, if not tens of millions of different resistors would need to exist to cover every value. So you just use something called "preferred values" with their resistance values posted on them instead.
Answer:
Theoretical moles of V are 1.6 moles
Explanation:
The theoretical yield of a reaction is defined as the amount of product you would make if all of the limiting reactant was converted into product.
In the reaction:
V2O5(s) + 5Ca(i) → 2V(i) + 5CaO(s)
Based on the reaction, 1 mol of V2O5 needs 5 moles of Ca for a complete reaction. As there are just 4 moles, <em>limiting reactant is Ca.</em> As there are produced 2 moles of V per 5mol of Ca, Theoretical moles of V are:
4 moles of Ca × (2mol V / 5Ca) = <em>1.6 moles of V</em>
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I hope it helps!
