Answer:
5.31x10⁻⁶ C
Explanation:
The cube is located 100 m altitude from the ground, so the superior face is at 100m and has E = 70 N/C, and the inferior face is at the ground with E = 130 N/C.
The electric field is perpendicular to the bottom and the top of the cube, so the total flux is the flux at the superior face plus the flux at the inferior face:
Фtotal = Ф100m + Фground
Where Ф = E*A*cos(α). α is the angle between the area vector and the field (180° at the topo and 0° at the bottom):
Фtotal = E100*A*cos(180°) + Eground*A*cos(0°)
Фtotal = 70A*(-1) + 130*A*1
Фtotal = 60A
By Gauss' Law, the flux is:
Фtotal = q/ε, where q is the charge, and ε is the permittivity constant in vacuum = 8.854x10⁻¹² C²/N.m²
A = 100mx100m = 10000 m²
q = 60*10000*8.854x10⁻¹²
q = 5.31x10⁻⁶ C
That would be the NOBLE GASES (Helium, Neon, Argon, Krypton, Xenon, Radon). Because these elements have a filled outer shell (thus giving them the full octet that other elements crave), they are stable elements under normal circumstances and as such resist chemical combination.
Plz note that under special conditions, noble gases such as Xenon and Radon can form compounds (Xenon Fluoride and Oxide; Radon Fluoride)
Answer:
Explanation:
I believe it is one because the energy is stored right before it falls
For this problem, we use the Hess' Law.
ΔHrxn = ∑(ν*Hf of products) - ∑(ν*Hf of reactants)
The ν represents the corresponding stoichiometric coefficients of the substances, while Hf is the heat of formation. For pure elements, Hf = 0.
Hf of Al₂O₃ = <span>−1676.4 kJ/mol
</span>Hf of Fe₂O₃ = <span>-826.0 kJ/mol
Thus,
</span>ΔHrxn = 1*−1676.4 kJ/mol + 1*-826.0 kJ/mol
<em>ΔHrxn = -2502.4 kJ/mol</em>
