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Otrada [13]
3 years ago
5

Suppose that the model presented by student 1 is correct. Based on the information provided, what would be the bond angle in a m

olecule of perchlorate ion
Physics
1 answer:
Ugo [173]3 years ago
8 0

Question: Predicting the shape of a molecule is relatively straight forward. A molecule's shape will always be determined by the number of electron pairs around the central atom. The number of electron pair corresponds to the number of atoms that are bound to the central atom of the molecule. For example, water contains two hydrogen atom bound to one atom of oxygen, giving the molecule a linear geometry.

Suppose that the model presented by student 1 is correct. Based on the information provided, what would be the bond angle in a molecule of perchlorate ion.

Answer: Suppose that the model presented by student 1 is correct The (perchlorate ion) will be a tetrahedral shape, O-Cl-O bond angle 109.5 due to four groups of bonding electrons and no lone pairs of electrons.

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Mv + mv = 2mv providing each momentum is in the same direction.
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if a big stone and a small stone is dropped simultaneously from the certain height. they fall together why??​
Alenkasestr [34]

Answer:

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A certain type of laser emits light that has a frequency of 4.2 × 1014 Hz. The light, however, occurs as a series of short pulse
bogdanovich [222]

Explanation:

It is given that,

Frequency of the laser light, f=4.2\times 10^{14}\ Hz

Time, t=3.2\times 10^{-11}\ s

(a) Let \lambda is the wavelength of this light. It can be calculated as :

\lambda=\dfrac{c}{f}

\lambda=\dfrac{3\times 10^8}{4.2\times 10^{14}}

\lambda=7.14\times 10^{-7}\ m

or

\lambda=714\ nm

(b) Let n is the number of the wavelengths in one pulse. It can be calculated as :

n=f\times t

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n = 13440

Hence, this is the required solution.

8 0
3 years ago
A basketball player is 4.22 m from
max2010maxim [7]

Answer: The height above the release point is 2.96 meters.

Explanation:

The acceleration of the ball is the gravitational acceleration in the y axis.

A = (0, -9.8m/s^)

For the velocity we can integrate over time and get:

V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))

for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)

P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)

now, the time at wich the horizontal displacement is 4.22 m will be:

4.22m = 9.20*cos(69°)*t

t = (4.22/ 9.20*cos(69°)) = 1.28s

Now we evaluate the y-position in this time:

h =  -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m

The height above the release point is 2.96 meters.

3 0
3 years ago
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A baseball is thrown with an initial velocity of 45.4 m/s at an angle of 31.2 ∘ .
sveticcg [70]

Answer:

V (initial vertical velocity) = 45.4 sin 31.2 = 23.52 m/s

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Check:

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8 0
2 years ago
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