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Lerok [7]
3 years ago
13

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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Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe
AlexFokin [52]

Answer:

0.56 atm

Explanation:

First of all, we need to find the number of moles of the gas.

We know that

m = 1.00 g is the mass of the gas

Mm=44.0 g/mol is the molar mass of the carbon dioxide

So, the number of moles of the gas is

n=\frac{m}{M_m}=\frac{1.00 g}{44.0 g/mol}=0.023 mol

Now we can find the pressure of the gas by using the ideal gas equation:

pV=nRT

where

p is the pressure

V=1.00 L = 0.001 m^3 is the volume

n = 0.023 mol is the number of moles

R=8.314 J/mol K is the gas constant

T=25.0^{\circ}+273=298 K is the temperature of the gas

Solving the equation for p, we find

p=\frac{nRT}{V}=\frac{(0.023 mol)(8.314 J/mol K)(298 K)}{0.001 m^3}=5.7 \cdot 10^4 Pa

And since we have

1 atm = 1.01\cdot 10^5 Pa

the pressure in atmospheres is

p=\frac{5.7\cdot 10^4 Pa}{1.01\cdot 10^5 Pa/atm}=0.56 atm

5 0
3 years ago
A balloon that has been rubbed on wool has a negative charge. When the balloon is held over—but not touching—some bits of paper,
alexandr402 [8]

Answer:

This process is known as static electricity

Explanation:

When balloon is rubbed on the wool , a negative charge is usually formed. When the balloon is then held over—but not touching—some bits of paper, some of the bits jump up to the balloon. The bits which jump up to the balloon are positive charged and we know that unlike charges attract.

This process explains static electricity which is when the charges on a material are not in a balanced state.

5 0
3 years ago
A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. its speed is 40 m/s when i
Alenkinab [10]

Given that,

Initial velocity , Vi = 0

Final velocity , Vf = 40 m/s

Acceleration due to gravity , a = 9.81 m/s²

Distance can be calculated as,

2as = Vf² - Vi²

2 * 9.81 *s = 40² - 0²

s = 81.55 m

For half height, that is, s = 40.77m

Vf= ??

2as = Vf² - Vi²

2 * 9.81 * 40.77 = Vf² - 0²

Vf² = 800

Vf = 28.28 m/s

Therefore, speed of roller coaster when height is half of its starting point will be 28 m/s.  

5 0
3 years ago
I need guidance pls
seropon [69]
The slowest line is the solid line and the fastest is the dotted line that crosses the solid line
for future reference you just need to find the slope or the line which is traveling most vertical
8 0
3 years ago
In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 48
cluponka [151]

Answer:

The thickness is  \Delta y =  2.4 *10^{-6} \  m

Explanation:

From the question we are told that

   The wavelength is  \lambda  = 480 \ nm = 480*10^{-9} \  m

    The first order of the dark  fringe is  m_1 =  16

     The second order of dark fringe considered is  m_2 = 6

Generally the condition for destructive interference is mathematically represented as

        y = \frac{m \lambda}{2}

Here y is the path difference between the central maxima(i.e the origin) and any dark fringe

So  the path difference between the 16th dark fringe and the 6th dark fringe is mathematically represented as

      y_1 - y_2 = \Delta y =  \frac{m_1 \lambda}{2} -  \frac{m_2 \lambda}{2}

=>  y_1 - y_2 = \Delta y =  \frac{16 *480*10^{-9}}{2} -  \frac{6 *480*10^{-9}}{2}

=>  y_1 - y_2 = \Delta y =  5 (480*10^{-9})

=>  \Delta y =  2.4 *10^{-6} \  m

8 0
3 years ago
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