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3 years ago

How many orbitals are allowed in a subshell if the angular momentum quantum number for electrons in that subshell is 3?

Physics

1 answer:

svp [43]3 years ago

4 0

Answer:

7 orbitals are allowed in a sub shell if the angular momentum quantum number for electrons in that sub shell is 3.

Explanation:

We that different values of m for a given l provide the total number of ways in which a given s, p,d and f sub shells in presence of magnetic field can be arranged in space along x, y ,z- axis or total number of orbitals into which a given subshell can be divided.

Range for given l lies between -l to +l .

The possible values of m are -3 , -2 , -1 , 0 , 1 ,2 , 3 .

Total number of orbitals are 7.

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Vlad1618 [11]

Answer:

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3 years ago

an empty density bottle weighed 23.5 and 48.4g when filled with water .The empty bottle was partially filled with sand until and

gavmur [86]

Answer:

Density of Sand is $2.653g/cm^{3}$ .

Explanation:

Given Empty Density bottle weighs 23.5gm(W=23.5gm)

Weight of bottle when completely filled water=48.4gm

So amount of water required to fill the bottle=Weight of bottle filled with water-W

Amount of water required to fill the bottle( $w_{max}$ )=48.4gm-23.5gm

$w_{max}=24.9$ g

Since we know density of water $d_{w} =1g/cm^{3}$ and $w_{max}$

We can calculate volume of empty space in the bottle(V).

$w_{max}=d_{w}$ $\times$ V

V= $\frac{w_{max} }{d_{w} }$

V= $\frac{24.9}{1}$

V=24.9 $g/cm^{3}$

Now bottle is partially filled with sand,and weight of bottle is ( $w_{s}$ )36.5gm

So,

Amount of sand added ( $m_{s}$ )=36.5-Weight of the bottle

$m_{s}$ =13g

After filling the bottle with water again,the weight of the bottle becomes ( $W_{2}$ =56.5g)

Therefore,

amount of water added to the bottle of sand in grams = $W_{2}$ -36.5gm

amount of water added =56.5g-36.5g

amount of water added =20g

As the density of water = 1g/ $cm^{3}$

Amount of water (in grams )=Volume of water occupied

20=volume of water added

Therfore volume of water added to the sand filled bottle( $V_{w}$ )=20 $cm^{3}$

As we know the total volume of the water bottle(V),

Volume of the sand occupied in the water bottle=V- $V_{w}$

$V_{s}$ =24.9g-20g

$V_{s}$ =4.9g

We know,

Density=Mass/Volume

Therefore,

density of sand = $\frac{m_{s} }{V_{s} }$

density of sand = $\frac{13}{4.9}$

density of sand = $2.653g/cm^{3}$

8 0

3 years ago

During the execution of a play, a football player carries the ball for a distance of 33 m in the direction 58° north of east. To

podryga [215]

Answer:28 m

Explanation:

Given

Direction is $58^{\circ}$ North of east i.e. $58 ^{\circ}$ with x axis

Also ball moved by 33 m

therefore its east component is 33cos58=17.48 m

Northward component $=33sin58=27.98 m\approx 28 m$

8 0

3 years ago

What are the measurements that are used to<br><br> describe movement?

Feliz [49]

Answer: The SI unit

Explanation: The SI units are used to describe the distance an object moves

3 0

3 years ago

A dart gun shoots a dart with an angle of 45' above horizontal During the upward part of the trajectory the gravitational accele

ludmilkaskok [199]

Answer:

4. Downward and its value is constant

Explanation:

As this is a case of projectile motion, we use the reference frame where upward direction to be positive for $y$ , and in the same way to be negative in the downward direction. On another hand, we have that gravity is always acting this means that gravitational acceleration g is directed downward constantly over the dart not only during the upward but also during the downward part of the trajectory. And it is ruled by the following equations.