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3 years ago

Give the approximate bond angle for a molecule with an octahedral shape.A) 109.5°B) 180°C) 120°D) 105°E) 90°

Physics

1 answer:

lbvjy [14]3 years ago

8 0

Answer:

90°

Explanation:

A molecule with an octahedral shape has a central atom bonded by six atoms or groups of atoms with no lone pairs of electron. The ideal bond angles is 90°. Examples of compounds with octahedral shape are sulfur hexafluoride SF₆ and molybdenum hexacarbonyl Mo(CO)₆

For a molecule with an octahedral shape:

Number of Electron Groups = 6

Electron-Group Geometry = octahedral

Number of Lone Pairs = 0

Valence Shell Electron Pair Repulsion (VSEPR) Notation = AX₆E₀ or AX₆

Ideal Bond Angles = 90°

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3 years ago

A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s,

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$S= \frac{1}{2} a t^2$
where a is the acceleration. Using the data of the problem, we can find a:
$a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2$
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a) power output at t=2.0 s
The velocity at t=2.0 s is
$v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J$

and so the power output is
$P= \frac{E}{t} = \frac{448 J}{2.0 s} =224 W$

b) power output at t=4.0s
The velocity at t=4.0 s is
$v(t)=at=(2.0 m/s^2)(4.0 s)=8.0 m/s$

the kinetic energy of the sprinter is
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and so the power output is
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c) Power output at t=6.0 s
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$v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(6.0 m/s)^2=4032 J$

and so the power output is
$P= \frac{E}{t} = \frac{4032 J}{6.0 s} =672 W$

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