What is the percent by mass of sodium in Na2SO4? total mass of element in compound molar mass of compound Use %Element x 100
I'm assuming you wanted this equation balanced? If you typed it correctly, it's already balanced for you. :)
Answer: Li is the reducing agentg and O is the oxidizing agent.
Explanation:
1) The oxidizing agent is the one that is reduced and the reducing agent is the one that is oxidized.
2) The given reaction is:
4Li(s) + O₂ (g) → 2 Li₂O(s)
3) Determine the oxidation states of each atom:
Li(s): oxidation state = 0 (since it is alone)
O₂ (g): oxidation state = 0 (since it is alone)
Li in Li₂O (s) +1
O in Li₂O -2
That because 2× (+1) - 2 = 0.
4) Determine the changes:
Li went from 0 to + 1, therefore it got oxidized and it is the reducing agent.
O went from 0 to - 2, therefore it got reduced and it is the oxidizing agent.
The central iodine atom in triiodide has sp3d hybridization.In triiodide anion, the central iodine atom has three equatorial lone pairs of electrons and the terminal iodines are bonded axially in a linear shape. Electrons in sp3d hybridization are arranged in trigonal bipyramidal symmetry.
Explanation:
When you draw the Lewis structure of this particle, you'll realize that the central I atom has a pair of bonds and three individual pairs of electrons. as a result of there are five things around that central I atom, it's<span> sp3d hybridized.
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The bonds during a gas<span> (CH4) molecule </span>are fashioned<span> by four separate </span>however<span> equivalent orbitals; </span>one<span> 2s and </span>3<span> 2p orbitals of the carbon </span>interbreed<span> into four sp3 orbitals. </span>within the<span> ammonia molecule (NH3), 2s and 2p orbitals </span>produce<span> four sp3hybrid orbitals, </span>one among that<span> is occupied by a lone </span>try<span> of electrons.</span><span>
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Answer:
Dos quintos al cuadrado dividido cinco medios a la menos 1.
Explanation:
