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Cloud [144]
3 years ago
11

Sodium (Na) and potassium (K) are in the same group in the periodic table. Sodium is in the 11th position. How many valence elec

trons does potassium have?
0

1

9

11
Physics
2 answers:
lutik1710 [3]3 years ago
6 0

<u>Answer:</u> The correct answer is 1.

<u>Explanation:</u>

Valence electrons are defined as the electrons that are present in the outermost shell of an atom.

Sodium is the 11th element in the periodic table. the number of electrons in this element are 11.

The electronic configuration of this element is 1s^22s^22p^63s^1

Valence shell of this element is 3s and number of electrons present in this shell is 1.

Hence, the correct answer is 1.

Nikolay [14]3 years ago
3 0

The correct answer is

1

Potassium has 1 valence electron. In fact, it is in the same group of sodium in the periodic table. But sodium has 1 valence electron, since it is in the 11th position, because the first two elements of the periodic table fills the first orbital, while the elements from 3 to 10 fill the second orbital (made of 8 electrons), so the 11th element (sodium) has 1 valence electron.

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What is the slope of the line if the rise of a line on a distance versus-time graph is 900 meters and the run is 3 minutes?
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6 0
3 years ago
the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

R=Range

R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

7 0
3 years ago
The magnitude of the weight of a 3.0 kg object on the surface of the earth is 29 N. True False
madreJ [45]
True

In fact, the weight of an object on the surface of the Earth is given by:
F=mg
where m is the mass of the object and g=9.81 m/s^2 is the gravitational acceleration on Earth's surface. If we use the mass of the object, m=3.0 kg, we find
F=mg=(3.0 kg)(9.81 m/s^2)=29 N
8 0
3 years ago
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