Pluto is a dwarf planet, but one of the largest known members, in the Kuiper belt.
The Kuiper Belt extends between 30 AU and 55 AU from the Sun
(1 AU = 1.5 × 10^8 km = distance from Earth to Sun).
Pluto's orbit is highly elliptical. It ranges from 30 AU to 50 AU. When Pluto is closest to the Sun, it is inside the orbit of Neptune (30 AU).
Astronomers class Pluto as a <em>resonant Kuiper belt object</em> (KBO). Because it gets so close to Neptune, its orbit is in <em>resonance</em> with that of Neptune. Pluto makes two orbits for every three of Neptune.
The number of moles of b2o3 that will be formed is determined as 4 moles.
<h3>
Limiting reagent</h3>
The limiting reagent is the reactant that will be completely used up.
4 b + 3O₂ → 2b₂O₃
from the equation above;
4 b ------------> 2 b₂O₃
2b ------------> b₂O₃
2 : 1
3O₂ -------------> 2b₂O₃
3 : 2
b is the limiting reagent, thus, the amount of b2o3 to be formed is calculated as;
4 b ------------> 2 moles of b2o3
8 moles -------> ?
= (8 x 2)/4
= 4 moles
Thus, the number of moles of b2o3 that will be formed is determined as 4 moles.
Learn more about limiting reactants here: brainly.com/question/14222359
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Answer:
The key processes of the rock cycle are crystallization, erosion and sedimentation, and metamorphism.
Explanation:
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This element is found in group 3A, period 3
<h3>Further explanation
</h3>
The maximum number of electrons that can be filled in the nth electron shell is 2n²(n=shell)
-
K shell (n = 1) maximum 2 x 1² = 2 electrons
- L shell (n = 2) maximum 2 x 2² = 8 electrons
- M shell (n = 3) maximum 2 x 3² = 18 electrons
- N shell (n = 4) maximum 2 x 4² = 32 electrons
Electron configuration of element X : 2.8.3 , so :
K shell = 2 ⇒1s²
L shell = 8⇒2s²2p⁶
M shell = 3⇒ 3s²3p¹
Block p: group 13-18 (has a 2p-6p configuration), also called a representative element because it includes metals, non-metals and metalloids
The outer shell 3s²3p¹ : located in group 3A and period 3
group⇒valence electron ⇒3
period⇒the greatest value of the quantum number n⇒3
Answer:
ΔHrxn = [(1) -1675.5 ( kJ/mole) + (2) 0 ( kJ/mole)] - [(1) -824.3 ( kJ/mole) + (2) 0 ( kJ/mole)]
Explanation:
ΔHrxn = 2ΔHf (Al₂O₃) - ΔHf (Fe₂O₃)
Remember that for pure elements in their standard state of temperature and pressure by definition their standard heats of formation are zero.
ΔHrxn = 2(-1675.7) - (-824.3) kJ/mol
ΔHrxn = 2527 kJ/mol
