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marta [7]
3 years ago
14

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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6 0
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}=24.9g

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}\timesV

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})=20cm^{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.

For the x-axis

v_{x}=v_{0}cos(45\°)=constant

x=(v_{0}cos(45\°))t

For the y-axis

v_{y}=v_{0}sin(45\°)-gt

y=v_{0}sin(45\°)t-\frac{1}{2} gt^{2}

Where v_{0}, is the initial velocity.

8 0
3 years ago
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