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\(5(1)^2 - 4(1) - 1 = 5 - 4 - 1 = 0\)
Remainder \(= 0\).
Escape velocity from Earth depends on:
v_esc = √(2GM/R), independent of mass of body and angle.
Using Gay-Lussac's Law at constant volume:
$\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$
$P_1 = 300\ \text{kPa} = 3 \times 10^5\ \text{Pa},\ T_1 = 300\ \text{K}$
$P_2 = 1.2 \times 10^6\ \text{Pa}$
$T_2 = T_1 \times \dfrac{P_2}{P_1} = 300 \times \dfrac{1.2\times10^6}{3\times10^5} = 300 \times 4 = 1200\ \text{K}$
$T_2\ (°C) = 1200 - 273 = \mathbf{927}°\text{C}$
- Cl < S: both in Period 3, S is to the left of Cl → S has larger radius ✓ → so Cl > S is wrong
- V > Ti: both 3d metals, generally decreases slightly left to right, but V > Ti can be considered close
- Rb < Cs (down Group 1, radius increases)
- Ne < Be (Ne has higher Z and is in Period 2)
4 g of Mg heated in excess O₂. Theoretical yield of MgO (g) is: \(2\mathrm{Mg} + \mathrm{O}_2 \rightarrow 2\mathrm{MgO}\)
Molar mass Mg = 24 g/mol: moles Mg = 4/24 = 0.1667 mol. 2 mol Mg → 2 mol MgO, so mol MgO = 0.1667 mol. Molar mass MgO = 40 g/mol: mass = 0.1667×40 = 6.67 g ≈ 6.6 g.
The product of upper-triangular matrices $\begin{pmatrix}1&k\\0&1\end{pmatrix}$ gives $\begin{pmatrix}1&\sum k\\0&1\end{pmatrix}$.
$\sum_{k=1}^{n-1}k = \dfrac{(n-1)n}{2} = 78 \Rightarrow n(n-1)=156 \Rightarrow n=13$
The inverse of $\begin{pmatrix}1&n\\0&1\end{pmatrix} = \begin{pmatrix}1&13\\0&1\end{pmatrix}$ is $\begin{pmatrix}1&-13\\0&1\end{pmatrix}$.
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