ΔG > 0
is always true for the freezing of water.
Explanation:
- The freezing of water is only spontaneous when the temperature is fairly small. Over 273 K, the higher value of TΔS causes the sign of ΔG to be positive, and there is no freezing point.
- The entropy decreases as water freezes. This does not infringe the Thermodynamics second law. The second law doesn't suggest entropy will never diminish anywhere.
- Entropy will decline elsewhere, provided it increases by at least as much elsewhere.
Answer:
300K.
Explanation:
The following data were obtained from the question:
Initial volume (V1) = 300L
Initial temperature (T1) = 200K
Final volume (V2) = 450L
Final temperature (T2) =..?
Since the pressure is constant, the gas is obeying Charles' law.
Using the Charles' law equation, we can obtain the new temperature of the gas as follow:
V1/T1 = V2/T2
300/200 = 450/T2
Cross multiply to express in linear form
300 x T2 = 200 x 450
Divide both side by 300
T2 = (200 x 450)/ 300
T2 = 300K
Therefore, the new temperature of the gas is 300K.
Ml=−2,−1,0,+1,+2.
<span>Since each of these orbitals can hold a maximum of </span>two electrons<span>, one having spin-up and one having spin-down, a total of </span>10 electrons<span> can share the quantum numbers n = 4 and l = 2</span>
The solution 550 ml total and first we will find the amount of alcohol. 3% = 0.03 550 ml x .03 = 16.5 ml alcohol
Then to find the amount of water used, we just have to subtract the amount of alcohol from the total volume
550 ml total - 16.5 ml alcohol = 533.5 ml water
Answer:
Dissolve 226 g of KCl in enough water to make 1.5 L of solution
Explanation:
1. Calculate the moles of KCl needed
2. Calculate the mass of KCl
3. Prepare the solution
- Measure out 224 g of KCl.
- Dissolve the KCl in a few hundred millilitres of distilled water.
- Add enough water to make 1.5 L of solution.
Mix thoroughly to get a uniform solution.
