Correct Answer: Option C
Reason:
<span>The </span>Pauli Exclusion Principle<span> states as '<em>in an atom or molecule, no two electrons can have the same four electronic quantum numbers. Further, an orbital can contain a maximum of only two electrons, the two electrons must have opposing spins.</em>'
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Thus, it can be seen that in option C, electrons in last 2 subshell have electrons with same spin, which is a violation of Pauli Exclusion Principle .
This family (ethane, propane, butane, etc) of materials is likely to have following set of properties.
- The alkanes are non- polar solvents.
- The alkanes are immiscible in water but freely miscible in other non-polar solvent .
- The alkanes are consisting of weak dipole dipole bonds can not breaks the strong hydrogen bond.
- The alkanes having only carbon (C) and hydrogen (H) atom which is bonded by a single bonds only.
- The alkanes posses weak force of attraction that is weak van der waals force of attraction.
The ethane, propane, butane, belong to alkanes family.The alkanes are also considers as saturated hudrocarbons. Ethane is found in gaseous stae Ethane is the second alkane followed by propane followed by butane.
learn about butane
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<span>4 Al + 3 O2 → 2 Al2O3
(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2
0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.
((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =
10.1 g O2 left over</span><span>
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