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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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El valor más preciso de la masa de un electron es 9.11*10^-11kg ¿ Cuánto aumentaría la masa de un cuerpo que se carga con -1 c d

chubhunter [2.5K]

Answer:

$m = 569.375\times 10^{6}\,kg$

Explanation:

Un electrón tiene una carga negativa de $1.6\times 10^{-19}\,C$ y la masa total es igual al producto del número de electrones y esa carga unitaria. El número de electrones se obtiene al dividir la carga total por la carga unitaria. (An electron has a negative charge of $1.6\times 10^{-19}\,C$ and the total mass is equal to the product of the qunatity of electrons and such unit charge. The quantity of electrons is found by diving the total charge by the unit charge):

$m = \frac{Q}{q} \cdot m_{e}$

$m = \left(\frac{1\,C}{1.6\times 10^{-19}\,C} \right)\cdot (9.11\times 10^{-11}\,kg)$

$m = 569.375\times 10^{6}\,kg$

7 0

3 years ago

PLEASE HELP ME I AM TIMED!

Crank

Answer:

violet has the highest frequency

Explanation:

7 0

3 years ago

In the great shopping cart race, two students push on shopping carts. A having twice the mass of B, with the same force applied

Lesechka [4]

K = 1/2 m x v^2

m = mass on the cart

V = velocity imparted to the cart

KA = 1/2 mA x vA^2.......................(1)

KB = 1/2 mB x vB^2........................(2)

Diving equation 1 by equation 2, we get -

KA/KB = mA/mB

= 2

KA = 2 x KB

Option A is correct

6 0

3 years ago

To lift a load of 100 kg a distance of 1 m an effort of 25 kg must be applied over an inclined plane of length 4 m. What must be

andreyandreev [35.5K]

Work = force * distance.
We must produce twice as much energy as we are lifting the weight twice as high.
But we are not increasing the force so we must increase the length of the ramp ( distance ) instead.
The new length will be twice as great as the previous length.
So 8 metres is required.

25 kg * 8 m = work = 100 kg * 2 m

7 0

3 years ago

A child on a tricycle is moving at a speed of 1.40 m/s at the start of a 2.25 m high and 12.4 m long incline. The total mass is

goblinko [34]

Answer:

The work done by the child as the tricycle travels down the incline is 416.96 J

Explanation:

Given;

initial velocity of the child, $v_i$ = 1.4 m/s

final velocity of the child, $v_f$ = 6.5 m/s

initial height of the inclined plane, h = 2.25 m

length of the inclined plane, L = 12.4 m

total mass, m = 48 kg

frictional force, $f_k$ = 41 N

The work done by the child is calculated as;

$\Delta E_{mech} = W - f_{k} \Delta L\\\\W = \Delta E_{mech} + f_{k} \Delta L\\\\W = (K.E_f - K.E_i) + (P.E_f - P.E_i) + f_{k} \Delta L\\\\W = \frac{1}{2} m(v_f^2 - v_i^2) + mg(h_f - h_i) + f_{k} \Delta L\\\\W = \frac{1}{2} \times 48(6.5^2 - 1.4^2) + 48\times 9.8(0-2.25) + (41\times 12.4)\\\\W = 966.96 \ - \ 1058.4 \ + \ 508.4\\\\W = 416.96 \ J$

Therefore, the work done by the child as the tricycle travels down the incline is 416.96 J

5 0

3 years ago

Why the substance on the periodic talbe classified together​

Why the substance on the periodic talbe classified together