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4 years ago

Where on the periodic table would this element most likely be found? Group 1 Group 2 Group 12 Group 13

Physics

1 answer:

andrezito [222]4 years ago

6 0

Answer:

The right approach will be Option C "Group 12".

Explanation:

A table that further arranges the chemical components in addition to expanding the electron configuration. Elements possessing identical characteristics are positioned in almost the same section (column or group), although elements in almost the same row become positioned with a similar amount of electron spheres.
Group 12 components therefore partially replaced (n − 1)d subshells, and therefore aren't metals, technically accurate. However, although much their chemistry seems to be comparable to something like the components throughout the d block which includes establishing themselves.

Some other options in question are not relevant to something like the given scenario. Because then the option here is just the right one.

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2 years ago

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Vinvika [58]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

A closed container initially holds 50 monatomic Aparticles that have a combined energy of 480 units. After 100 monatomic B parti

Molodets [167]

Answer:

"8 units" is the appropriate answer.

Explanation:

According to the question,

Throughout equilibrium all particles are of equivalent intensity, and as such the integrated platform's total energy has been uniformly divided across all individuals.

Now,

The total energy will be:

= $480+720$

= $1200 \ units$

The total number of particles will be:

= $50+100$

= $150$

hence,

Energy of each A particle or each B particle will be:

= $\frac{1200}{150}$

= $8 \ units$

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3 years ago

10. A change in

pogonyaev

Answer:

d velocity will be the one according to me

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3 years ago

At noon, ship A is 110 km west of ship B. Ship A is sailing east at 20 km/h and ship B is sailing north at 15 km/h. How fast is

Katyanochek1 [597]

Answer:

$4.47\ \text{km/h}$

Explanation:

$\dfrac{da}{dt}$ = Rate at which the distance between A and starting point of B is changing = -20 km/h

$\dfrac{db}{dt}$ = Rate at which the distance of B is changing = 15 km/h

$\dfrac{dc}{dt}$ = Rate at which the distance between A and B is changing

Time after which the rate at which the distance between A and B is changing is 4 hours

Distance covered by A in 4 hours = $20\times 4=80\ \text{km}$

a = Distance remaining to the start point of B = $110-80=30\ \text{km}$

b = Distance covered by B in 4 hours = $15\times 4=60\ \text{km}$

Distance between A and B after 4 hours

$c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}$

$c^2=a^2+b^2$

Differentiating with respect to time we get

$c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}$

The rate at which the distance between the ships is changing at 4 PM is $4.47\ \text{km/h}$ .

7 0

3 years ago