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3 years ago

When four atomic orbitals are mixed to form hybrid orbitals, how many hybrid orbitals are formed?

Physics

2 answers:

lilavasa [31]3 years ago

6 0

When four atomic orbitals are mixed to form hybrid orbitals, 4 x sp3 hybrid orbitals are formed.

Carbon's 2s and all three of its 3p orbitals hybridize to form four sp3 orbitals. These orbitals then bond with four hydrogen atoms through sp3-s orbital overlap, creating methane.

Snezhnost [94]3 years ago

5 0

Answer:

Four

Explanation:

When atomic orbitals are combined, hybrid orbitals are formed. Hybrid orbitals exist only in covalent-bonded atoms and not in isolated atoms. The shapes and orientations of hybrids are different from that of atomic orbitals. The number of hybrid orbitals is same as the number of atomic orbitals that combined to form atomic orbitals. Thus, when 4 atomic orbitals are mixed then 4 hybrid orbitals are formed.

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Answer:

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Explanation:

information we know:

Total force: $F=45N$

Weight: $w=100N$

distance: $4m$

vertical component of the force: $F_{y}=12N$

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: $F_{x}=Fcos\theta$

vertical component: $F_{y}=Fsen\theta$

but from the given information we know that $F_{y}=12N$

so, equation these two $F_{y}=Fsen\theta$ and $F_{y}=12N$

$Fsen\theta =12N$

and we know the force $F=45N$ , thus:

$45sen\theta=12$

now we clear for $\theta$

$sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466$

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

$F_{x}=Fcos\theta$

$F_{x}=45cos(15.466)\\F_{x}=43.37N$

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

$W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J$

the work done is W=173.48J

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Explanation:

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