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What mass of \(\mathrm{Al}_2\mathrm{O}_3\) is produced from 18.5 g Al? \(4\mathrm{Al} + 3\mathrm{O}_2 \rightarrow 2\mathrm{Al}_2\mathrm{O}_3\)
Molar mass Al = 27 g/mol: moles Al = 18.5/27 = 0.685 mol. From stoichiometry: 4 mol Al → 2 mol Al₂O₃, so mol Al₂O₃ = 0.685/2 = 0.3425 mol. Molar mass Al₂O₃ = 102 g/mol: mass = 0.3425×102 = 34.94 g ≈ 34.9 g.
Let $S$ be the set of distinct values of $b$ for which the linear system $x+y+z=1$, $x+ay+z=1$, $ax+by+z=0$ has no solution. Describe the set $S$.
Subtract equation 1 from equation 2: $(a-1)y = 0$.
Case $a\neq1$: $y=0$. Then from eq.1 and eq.3: $x+z=1$ and $ax+z=0$, giving $(a-1)x=-1$, so $x=\frac{-1}{a-1}$. This always has a solution for $a\neq1$.
Case $a=1$: System becomes $x+y+z=1$, $x+y+z=1$, $x+by+z=0$. The third equation forces $-1=by-... $ i.e. subtracting eq.1: $(b-1)y=-1$. For no solution we need $b=1$.
So $S=\{1\}$ — a single value. Hence $S$ is a singleton.
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