A tetrahedral shape corresponds to sp3 hybridization.

In CH4, the central carbon atom forms four equivalent bonds with hydrogen atoms, indicating sp3 hybridization.

The carbon atom in CO2 is sp hybridized, forming two double bonds with oxygen atoms.

Oxygen in H2O is sp3 hybridized, forming two bonds and two lone pairs.

The central nitrogen atom in N2O is sp hybridized, leading to a linear arrangement.

The central nitrogen atom in NH3 is sp3 hybridized due to three bonding pairs and one lone pair.

The central atom nitrogen in NH4+ undergoes sp3 hybridization to form four equivalent bonds with hydrogen.

The central oxygen atom in O3 is sp2 hybridized, leading to a bent molecular shape.

In ozone (O3), the central oxygen atom is sp2 hybridized, resulting in a bent molecular shape.

In PCl5, phosphorus is sp3d hybridized, allowing for five bonding pairs.

The central sulfur atom in SF6 is surrounded by six fluorine atoms, leading to an octahedral geometry and sp3d2 hybridization.

The sulfur atom in SO2 is sp2 hybridized, forming one double bond and one single bond.

The xenon atom in XeF4 is dsp3 hybridized, resulting in a square planar geometry.

In [Ni(CN)4]2-, nickel is in the +2 oxidation state and undergoes dsp2 hybridization to accommodate the four cyanide ligands.

In [Ni(CO)4], the nickel atom undergoes sp3 hybridization to accommodate four CO ligands.

In [Cu(NH3)4]2+, the copper ion undergoes dsp2 hybridization to accommodate four ligands.

In [Cr(NH3)6]3+, the hybridization is d2sp3 due to the octahedral arrangement of six ligands.

In [Ni(CO)4], the nickel ion undergoes sp3 hybridization to accommodate four CO ligands.

The nitrogen atom in a primary amine is sp3 hybridized.

The nitrogen atom in amines is sp3 hybridized, as it forms three sigma bonds and has one lone pair.

The nitrogen atom in amines is sp3 hybridized, having a tetrahedral geometry with one lone pair.

In NH3, nitrogen has three bonds and one lone pair, leading to sp3 hybridization.

In PCl5, phosphorus is surrounded by five chlorine atoms, indicating d2sp3 hybridization.

Hydrostatic pressure is given by P = ρgh. Thus, P = 1000 kg/m³ * 9.81 m/s² * 10 m = 98100 Pa.

The ideal gas equation is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Using the lens formula, 1/f = 1/v - 1/u, we find v = -8 cm.

Using the lens formula 1/f = 1/v - 1/u, we find v = 15 cm.

Using the lens formula 1/f = 1/v - 1/u, we find v = 15 cm.

Using the lens formula 1/f = 1/v - 1/u, we find v = 20 cm.

Using the lens formula 1/f = 1/v - 1/u, we have 1/10 = 1/v - 1/(-30). Solving gives v = 15 cm.