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frutty [35]
3 years ago
11

What is the difference between the number of electrons in an atom of selenium, Se, and the number of electrons in an atom of alu

minum, Al?
Physics
2 answers:
Bumek [7]3 years ago
8 0

Answer : The difference is that the selenium has more number of electrons as compared to aluminum.

Explanation :

The difference between the number of electrons in an atom of selenium, Se, and the number of electrons in an atom of aluminum, Al are:

As we know that,

Element selenium has the atomic number 34 while the aluminium has atomic number of 13.

Element selenium belongs to the group 16 while aluminium belongs to group 3. The group determine the number of electron in its outermost shell.

This means that, selenium has six electrons in its outermost shell and aluminium has only three electrons in its outermost shell.

The electronic configuration of selenium is, 1s^22s^22p^63s^23p^63d^{10}4s^24p^4 while electronic configuration of aluminium is, 1s^22s^22p^63s^23p^1

Thus, the difference between the number of electrons in an atom of selenium and the number of electrons in an atom of aluminum is that the selenium has more number of electrons as compared to aluminum.

Oxana [17]3 years ago
4 0
Well, electrons can be converted into a atomic number so if SE atomic number is 34 that means it has 34 electrons. AI has a atomic number of 13 meaning it has 13 electrons.

So the difference is that SE has more electrons then AI.

Hope this helped. :D
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Answer:

0.25223 seconds.

Explanation:

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m_2 = Mass of block = 2.55 kg

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g = Acceleration due to gravity = 9.81 m/s²

As linear momentum is conserved

m_1u_1 + m_2u_2 =(m_1 + m_2)v

Now u_2=v as the block (with the bullet in it) reverses direction and rises,

m_1u_1 + m_2v =(m_1 + m_2)v\\\Rightarrow m_1u_1=(m_1 + m_2)v-m_2v\\\Rightarrow 0.0146\times 816=(0.0146 + 2.55)v-(-2.55v)\\\Rightarrow 11.9136=2.2646v+2.55v\\\Rightarrow 11.9136=4.8146v\\\Rightarrow v=\frac{11.9136}{4.8146}\\\Rightarrow v=2.47447\ m/s

Equation of motion

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{2.47447-0}{9.81}\\\Rightarrow t=0.25223\ s

The time t is 0.25223 seconds.

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With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you releas
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Answer:

The initial velocity is 0.5114 m/s or 511.4 mm/s

Explanation:

Let the initial velocity be 'v'.

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Mass of the ball (m) = 130 g = 0.130 kg   [ 1 g = 0.001 kg]

Initial height of the ball (h₁) = 1.4 mm = 0.0014 m   [ 1 mm = 0.001 m]

Final height of the ball (h₂) = 15 mm = 0.015 m

Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.

Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.

Change in kinetic energy is given as:

\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity

As it just touches the 15 mm high roof, the final velocity will be zero. So,

v_f=0\ m/s.

Now, the change in kinetic energy is equal to:

\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2

Change in gravitational potential energy = Final PE - Initial PE

So,

\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J                    [ g = 9.8 m/s²]

Now, Change in KE = Change in PE

0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s

Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s

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