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:
Empirical formula: BH3
Molecular Formula: B2H6
Explanation:
To solve the exercise, we need to know how many boron atoms and how many hydrogen atoms the compound has. We know that of the total weight of the compound, 78.14% correspond to boron and 21.86% to hydrogen. As the weight of the compound is between 27 g and 28 g, using the above percentages we can solve that the compound has between 21.1 g and 21.8 g of boron, and between 5.9 g and 6.1 g of hydrogen:
100% _____ 27 g
78.14% _____ x = 78.14% * 27g / 100% = 21.1 g boron
100% ______27 g
21.86% ______ x = 21.86% * 27g / 100% = 5.9 g hydrogen
100% _____ 28 g
78.14% _____ x = 78.14% * 28g / 100% = 21.8 g boron
100% _____ 28g
21.86% _____ x = 21.86% * 28g / 100% = 6.1 g hydrogen
So, if the atomic weight of boron is 10.8 g, there must be two boron atoms in the compound that sum 21.6 g. The weight of hydrogen is 1 g, so the compound must have six hydrogen atoms.
The molecular formula represents the real amount of atoms that form a compound. Therefore, the molecular formula of the compound is B2H6.
The empirical formula is the minimum expression that represents the proportion of atoms in a compound. For example, ethane has 2 carbon atoms and 6 hydrogen atoms, so its molecular formula is C2H6, however, its empirical formula is CH3. Therefore, the empirical formula of the boron compound is BH3.
<span>2 NH</span>₃<span> + 3 O</span>₂<span> + 2 CH</span>₄<span> </span>⇒<span> 2 HCN + 6 H</span>₂<span>O
2mol : 2mol
34g : 54g
25,1g : x
x = (25,1g * 54g) / 34g </span>≈ 39,9g<span>
</span>
The symbol for xenon (xe) would be a part of the noble gas notation for the element cesium.
For writing the electronic configuration of any element by using the preceding noble gas configuration, we simply use the symbols of noble gas belongs to the previous period of that particular elements. We can't use the symbol of noble gas of same period from which the element belong.
A is the wrong option because the noble gas in the preceding period to the period from which antimony belongs is krypton.
The actual electronic configuration of antimony is as follow:
[Kr] 4d10 5s2 5p3
B is correct option because the noble gas in the preceding period to the period from which Cesium belongs is Xenon.
The actual electronic configuration of Cesium is as follow:
[Xe] 6s1
Thus, we concluded that the symbol for xenon (xe) would be a part of the noble gas notation for the element cesium.
learn more about Noble gas:
brainly.com/question/2094768
#SPJ4
It would definitely be hand lense
