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Fittoniya [83]
3 years ago
9

Valence electron of the first 20 elements​

Physics
2 answers:
Dima020 [189]3 years ago
4 0

Answer:

Hydrogen

1 valence electron

Helium

2 valence electrons

lithium

1 valence electrons

beryllium

2 valence electrons

boron

3 valence electrons

carbon

4 valence electrons

nitrogen

5 valence electrons

oxygen

6 valence electrons

flourine

7 valence electrons

neon

8 valence electrons

sodium

1 valence electron

magnesium

2 valence electrons

aluminum

3 valence electrons

silicon

4 valence electrons

phosphorus

5 valence electrons

bixtya [17]3 years ago
4 0

Answer:  17 Chlorine -1, +1, (+2), +3, (+4), +5, +7

18 Argon 0

19 Potassium +1

20 Calcium +2

Explanation:

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Can anyone explain and get the answer pls?
Veronika [31]

FL₁ =Fl₂

80.x₁ = 30.7

x₁=2.625 m

6 0
3 years ago
A current of 3.60A flows for 15.3s through a conductor. Calculate the number of electrons that pass through a point in the condu
Artyom0805 [142]

3.60 A = 3.60 coulombs of charge per second

(3.60 Coul/sec) x (15.3 sec) = 55.08 coulombs of charge

1 coulomb of charge is carried by 6.25 x 10^18 electrons

Number of electrons =

               (55.08 Coul) x (6.25 x 10^18 e/coul) = <em>3.4425 x 10^20 electrons</em>


7 0
4 years ago
Please I need help........
LenKa [72]

Answer:

7.46 J/kg/K

Explanation:

The heat absorbed or lost is:

q = mCΔT

where m is the mass, C is the heat capacity, and ΔT is the change in temperature.

Given q = 15.0 J, m = 0.201 kg, and ΔT = 10.0 °C:

15.0 J = (0.201 kg) C (10.0 °C)

C = 7.46 J/kg/°C

Which is the same as 7.46 J/kg/K.

7 0
3 years ago
Read 2 more answers
An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if
maxonik [38]
The electron is accelerated through a potential difference of \Delta V=780 V, so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
\frac{1}{2}mv^2 =  e \Delta V
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
\Delta V is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s


Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
evB=m \frac{v^2}{r}
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T
3 0
4 years ago
A bar magnet whose magnetic dipole moment is 21 A·m2 is aligned with an applied magnetic field of 3.6 T. How much work must you
iren [92.7K]

Answer:

The amount of work must be do to rotate the bar magnet is 151.2 J

Explanation:

Given:

Magnetic moment \mu = 21 A. m^{2}

Magnetic field B = 3.6 T

To find work do to rotate the bar magnet,

From the formula of work done in case of magnetic field,

    U = \mu .B \cos 0 -\mu .B \cos 180

Here \theta changes 0 to 180

But \cos 180 = -1

    U = 2\mu B

    U = 2 \times 21 \times 3.6

    U = 151.2 J

Therefore, the amount of work must be do to rotate the bar magnet is 151.2 J

3 0
3 years ago
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