Answer:
ΔH = 125.94kJ
Explanation:
It is possible to make algebraic sum of reactions to obtain ΔH of reactions (Hess's law). In the problem:
1. 2W(s) + 3O2(g) → 2WO3(s) ΔH = -1685.4 kJ
2. 2H2(g) + O2(g) → 2H2O(g) ΔH = -477.84 kJ
-1/2 (1):
WO3(s) → W(s) + 3/2O2(g) ΔH = 842.7kJ
3/2 (2):
3H2(g) + 3/2O2(g) → 3H2O(g) ΔH = -716.76kJ
The sum of last both reactions:
WO3(s) + 3H2(g) → W(s) + 3H2O(g)
ΔH = 842.7kJ -716.76kJ
<h3>ΔH = 125.94kJ </h3>
You would expect to see a marble-looking rock
<span>Rules for the Principal Energy Level. A principal energy level may contain up to 2n2 electrons, with n being the number of each level. The set of three p orbitals can hold up to 6 electrons. Thus, the second principal energy level can hold up to 8 electrons, 2 in the s orbital and 6 in the p orbital.</span>
Answer:
A. H2O
Explanation:
Let us first define the three types of bonds:
1. Nonpolar Covalent: electronegativity difference < 0.4
2. Polar Covalent: electronegativity difference between 0.4 and 1.8
3. Ionic: electronegativity difference > 1.8
This will help us eliminate choices C and D:
-NaCl has a electronegativity difference of 3.0 - 0.9 = 2.1 (ionic bond)
-Cl2 has a electronegativity difference of 3.0 - 3.0 = 0 (nonpolar covalent bond)
However, we still have two more options, A and B, but they are not diatomic for us to use the electronegativity differences with.
We must now consult their geometries. Because CO2 has a linear geometry (O=C=O), the two sides will cancel each other out, resulting in a nonpolar covalent bond. At this point, by process of elimination, we can already determine the answer to be A. H2O. We can verify this by looking at the geometry of H2O, which is bent (H-O-H; imagine the O is above the H's, I cannot draw it in this response). H2O's bent geometry classifies it as polar covalent; the electrons are slightly more attracted towards the O, the more electronegative element. Side note: this makes the O slightly more negative in charge, whilst the H's are slightly more positive in charge.
P.S. I apologize for not being able to draw and demonstrate that last paragraph, but I hope you get a general idea. You can search up the "H2O geometry" and "CO2 geometry" to get a better idea! :)
