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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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Assuming there are no accidents or delays, the distance that a car travels down the interstate
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Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.
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The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

m_{T}=\frac{F_{pull}}{a}

m_{T}=\frac{3615}{1.516}

m_{T}=2384.5

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

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