<span>XY4Z2-->Square planar (Electron domain geometry: Octahedral) sp3d2
XY4Z-->Seesaw (Electron domain geometry: Trigonal bipyramidal) sp3d
XY5Z-->Square pyramidal (Electron domain geometry: Octahedral) sp3d2
XY2Z3-->Linear (Electron domain geometry: Trigonal bipyramidal) sp3d
XY2Z-->Bent (Electron domain geometry: Trigonal planar) sp2
XY3Z-->Trigonal pyramidal (Electron domain geometry: Tetrahedral) sp3
XY2Z2-->Linear (Electron domain geometry: Tetrahedral) sp3
XY3Z2-->T shaped (Electron domain geometry: Trigonal bipryamidal) sp3d
XY2-->Linear (Electron domain geometry: Linear) sp
XY3 Trigonal planar (Electron geometry: Trigonal planar) sp2
XY4-->Tetrahedral (Electron domain geometry: tetrahedral) sp3
XY5-->Trigonal bipyramidal (Electron domain geometry: Trigonal bipyramidal) sp3d
XY6-->Octahedral (Electron domain geometry: Octahedral) sp3d2</span>
Answer:
2.6 kJ
Explanation:
The formula for the amount of heat (q) absorbed by the water is
q = mCΔT
1. Calculate ΔT
ΔT = 23.5 °C - 22.1 °C = 1.4 °C
2. Calculate q
q₂ = mCΔT = 500 g × 4.184 J·°C⁻¹g⁻¹ × 1.4 °C = 2900 J = 2.9 kJ
(1) MO₂(s) + C(s) → M(s) + CO₂ (g), ΔG₁ = 288.9 kJ/mol
(2) C(s) + O₂(g) → CO₂(g), ΔG₂ = -394.4 kJ/mol
By adding both equations 1 + 2 we get the coupled reaction:
MO₂(s) + 2 C(s) + O₂(g) → M(s) + 2 CO₂(g)
ΔG⁰ = ΔG₁ + ΔG₂
= 288.9 + (-394.4) = -105.5 kJ/mol = -105500 J/mol
Temperature T = 25 + 273.15 = 298.15 K
Molar gas constant R = 8.314 J/mol.K
K =
=
= 3.05 x 10¹⁸
Covalent network. <span>A solid that is extremely hard, that has a very high melting point, and that will not conduct electricity either as a solid or when molten is held together by a continuous three-dimensional network of covalent bonds. Examples include diamond, quartz (SiO </span><span>2 </span>), and silicon carbide (SiC). The electrons are constrained in pairs to a region on a line between the centers of pairs of atoms.<span>
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