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3 years ago

Why the substance on the periodic talbe classified together

Physics

2 answers:

denpristay [2]3 years ago

6 0

you mean table*, but idk

Thepotemich [5.8K]3 years ago

5 0

Hey there!

Here is your answer for this question!

Answer : The elements in a group look and behave similary because they both have the same number of Electrons as their outermost protective shell. Also often said as the face they show to the world around us. Group number 18 elements as on the far side to the right of the Periodic Table, for example, you have completely full outer shells and hardly participate in the chemical reactions you might see in a substance.

The groups of the periodic table of groups 1,2, 11,15,16,17 and 18 are down below!

Group #1. Alkali Metals.

Group #2. Alkaline Earth Metals.

Group #11. Coinage Metals (it is not an IUPAC approved as the name!)

Group #15. Pnictogens (also not used as an IUPAC approved name!)

Group #16. Chalcogens.

Group #17. Halogens.

Group #18. Noble Gases.

I hope this helps you! Let me know if you need anything else! ☺

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A 59 kg physics student is riding her 220 kg Harley at 12 m/s when she has a head-on collision with a 2.1 kg pigeon flying the o

IgorLugansk [536]

Answer:

Explanation:

Using conservation of momentum

$m_1=59+220=279kg$

$m_2=2.1kg$

$U_1=12m/s$

$U_2=-44m/s$

$m_1U_1+m_2U_2=(m_1+m_2)V\\\\\frac{278\times 12-2.1\times 44}{279+2.1}=V\\\\V=11.58m/s$

6 0

3 years ago

The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, what should be the focal length an

yuradex [85]

Answer:

The focal length of the appropriate corrective lens is 35.71 cm.

The power of the appropriate corrective lens is 0.028 D.

Explanation:

The expression for the lens formula is as follows;

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

Here, f is the focal length, u is the object distance and v is the image distance.

It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.

Put v= -71.4 cm and u= 24.0 cm in the above expression.

$\frac{1}{f}=\frac{1}{24}+\frac{1}{-71.4}$

$\frac{1}{f}=0.028$

f= 35.71 cm

Therefore, the focal length of the corrective lens is 35.71 cm.

The expression for the power of the lens is as follows;

$p=\frac{1}{f}$

Here, p is the power of the lens.

Put f= 35.71 cm.

$p=\frac{1}{35.71}$

p=0.028 D

Therefore, the power of the corrective lens is 0.028 D.

3 0

3 years ago

If the final speed of an object as is strikes the ground is 77 m/s and it was in the air for 6.5 seconds. What was the initial d

azamat

Answer:

The answer to your question is: 13.2 m/s

Explanation:

final speed (fs) = 77 m/s

t = 6.5 s

gravity (g) = 9.81 m/s2

initial speed (is) = ?

Formula

fs = is + gt from this equation we clear "is" = fs - gt

Substitution is = 77 - (9,81)(6.5)

Process is = 77 - 63.8

is = 13.2 m/s

8 0

3 years ago

A plane has an airspeed of 142 m/s. A 30.0 m/s wind is blowing southward at the same time as the plane is flying. What must be t

const2013 [10]

Answer:

$\theta=12.19^{\circ}$

Explanation:

Given that

The speed of the airplane ,v= 142 m/s

The speed of the air ,u = 30 m/s

Lets take angle make by airplane from east direction towards north direction is θ .

Now by using diagram ,we can say that

$sin\theta =\dfrac{u}{v}$

Now by putting the values in the above equation we get

$sin\theta =\dfrac{30}{142}$

$sin\theta=0.21$

$\theta=12.19^{\circ}$

Therefore the angle will be 12.19° .

4 0

3 years ago

The force of attraction between a -130.0 C and +180.0 C charge is 8.00 N. What is the separation between these two charges in me

kondaur [170]

Answer with Explanation:

The force of attraction between 2 charges of magnitude $q_1,q_2$ separated by a distance 'r' is given by

$F=\frac{1}{4\pi \epsilon _o}\frac{q_1\times q_2}{r^2}=k\frac{q_1\times q_2}{r^2}$

where

$\epsilon _o$ is a constant known as permitivity of free space

$k=9\times 10^{9}Nm^2/C^2$

Applying the given values in the above relation we get

$8=9\times 10^{9}\times \frac{130\times 180}{r^{2}}\\\\r^{2}=\frac{9\times 10^{9}\times 130\times 180}{8}\\\\r^{2}=26325\times 10^{9}\\\\\therefore r=\sqrt{26325\times 10^{9}}=5.131\times 10^{6}meters$

5 0

3 years ago

Why the substance on the periodic talbe classified together​

Why the substance on the periodic talbe classified together