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3 years ago

Which hybridization schemes allows the central atom to form more than four bonds?

Physics

2 answers:

Dmitry [639]3 years ago

7 0

The options in this question are sp3, sp3d and sp2

Answer:

sp3d

Explanation:

The hybridization scheme in a molecule determines the number of bonds the central atom can form with other atoms.

sp3 hybridization leads to the formation of four bonds by the central atom

sp2 hybridization leads to the formation of three bonds by the central atom

sp3d hybridization leads to the formation of five bonds by the central atom. Hence the answer.

olganol [36]3 years ago

5 0

Answer:

sp3d

Explanation:

sp=2

sp^2=3

sp^3=4

sp^3d=5

As we add, we can make one more bond...since this says MORE than four bond, the answer must be sp^3d.

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3 years ago

A train at a constant 60.0 km/h moves east for 40.0 min, then in a direction 50.0 degrees east of due north for 20.0 min, and th

son4ous [18]

Answer:

Part a)

$v = 7.57 km/h$

Part b)

$\theta = 67.5 degree$ North of East

Explanation:

Speed of train towards East = 60 km/h

displacement towards East is given as

$d_1 = 40 km$

now it turns towards 50 degree East of North

so its distance is given as

$d_2 = 20 km(sin50 \hat i + cos50\hat j)$

$d_2 = 15.3 \hat i + 12.8 \hat j$

then finally it moves towards west for 50 min

$d_3 = -50 \hat i$

Now the total displacement of the train is given as

$d = d_1 + d_2 + d_3$

$d = (40 + 15.3 - 50)\hat i + 12.8 \hat j$

$d = 5.3\hat i + 12.8 \hat j$

now total time duration of the motion is given as

$T = 40 min + 20 min + 50 min$

$T = 1.83 h$

now average velocity is given as

$v_{avg} = \frac{5.3\hat i + 12.8\hat j}{1.83}$

$v_{avg} = 2.89\hat i + 6.99\hat j$

Part a)

magnitude of the average velocity is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{2.89^2 + 6.99^2}$

$v = 7.57 km/h$

Part b)

Direction of the velocity is given as

$tan\theta = \frac{v_y}{v_x}$

$tan\theta = \frac{6.99}{2.89}$

$\theta = 67.5 degree$ North of East

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3 years ago

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Bumek [7]

Explanation:

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Hence, $\Delta E$ = 0. And, as $\Delta E$ = 0 then there will be no work involved. This means that total energy added to the house will return to the outside air as heat.

Therefore,

Q = -(19000 J + 2000 J)

= -21000 J

or, |Q| = 21000 J

Thus, we can conclude that the magnitude of the energy transfer between the house and the outside air is 21000 J.

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3 years ago

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alukav5142 [94]

Answer:

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