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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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v of the girl = 4.0 m/s

A) Momentum of the girl as she is diving:
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B) momentum of the raft = - momentum of the girl = -24.0 N/s

C) speed of the raft

p = m.v ; v = p/m = 24.0N/s / 180 kg = -0.13 m/s [i.e. in the opposite direction of the girl's velocity]

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An ideal gas Carnot cycle with air in a piston cylinder has a high temperature of 1200 K and a heat rejection at 400 K. During t

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Answer:

The specific heat capacity is q_{L}=126.12kJ/kg

The efficiency of the temperature is n_{TH}=0.67

Explanation:

The p-v diagram illustration is in the attachment

T_{H} means high temperature

T_{L} means low temperature

The energy equation :

$q_{h}$ = R* $T_{h}$ in( $V_{2}$ / $V_{1}$ )

$=0.287 * 1200 ln(3)$

$=0.287*1318.33$

$=378.36kJ/kg$

The specific heat capacity:

$q_{L}$ =q_{h}*(T_{L}/T_{H})

q_{L}=378.36 * (400/1200)

q_{L}=378.36 * 0.333

q_{L}=126.12kJ/kg

The efficiency of the temperature will be:

$n_{TH}$ =1 - ( $T_{L}$ / $T_{H}$ )

n_{TH}=1-(400/1200)

n_{TH}=1-0.333

n_{TH}=0.67

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A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A

Sphinxa [80]

Answer:

Density of 127 I = $\rm 1.79\times 10^{14}\ g/cm^3.$

Also, $\rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

Explanation:

Given, the radius of a nucleus is given as

$\rm r=kA^{1/3}$ .

where,

$\rm k = 1.3\times 10^{-13} cm.$

A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

$\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.$

For the nucleus 127 I,

Mass, M = $\rm 2.1\times 10^{-22}\ g.$

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

$\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.$

On comparing with the density of the solid iodine,

$\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

7 0

3 years ago

Why the substance on the periodic talbe classified together​

Why the substance on the periodic talbe classified together