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irina [24]
3 years ago
12

A scientist performs an experiment on an unknown element. She finds that the element forms an ionic bond with beryllium (Be) but

not with lithium (Li). She concludes that the unknown element must belong in group 2 of the periodic table. State whether or not you think this result supports her conclusion, and why
Physics
2 answers:
Bumek [7]3 years ago
4 0

To fill up all of the valence electrons we must have 8 electrons total to have an ionic bond.  In this case Lithium is in group 1 and so only has 1 valence electron and so it can bond with elements from group 17, since 7 + 1 = 8.  However, this unknown element did bond with beryllium, and beryllium is in group 2 so it must bond with an atom that has 6 valence electrons.  The elements that have 6 valence electrons are in group 16 not group 2.  So the result does not support her conclusion.

aivan3 [116]3 years ago
3 0

Answer: The result do not support her conclusion as group 2 elements are metals.

Explanation:

An ionic bond is formed when an element completely transfers its valence electron to another element. The element which donates the electron is known as electropositive element or the metal and forms a positively charged ion called as cation. The element which accepts the electrons is known as electronegative element or non metal and forms a negatively charged ion called as anion.

As both Lithium (Li) and berrylium (Be) are metals, they would combine with a non metal to form ionic bond. Thus the unknown element must be a non metal and should belong to group 14 , 15 , 16 or 17 and not to group 2, which are metals.

Thus The result do not support her conclusion as group 2 elements are metals.

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An inductor in an LC circuit has a maximum current of 2.4 A and a maximum energy of 56 mJ.
Harrizon [31]

Answer:

The energy stored in the capacitor, when the current in the inductor is 1.2 A, is 41.6 mJ.

Explanation:

In a LC oscillating circuit, the energy is stored in the electric field (between the plates of the capacitor) and in the magnetic field (surrounding the wires of the inductor).

At any time, the sum of both energies can be expressed as follows:

E = 1/2 Q² / C   +  1/2 L I²

In this type of circuit, energy oscillates, which means that it is exchanging between both fields all time.

When the capacitor is completely discharged, all the energy is stored in the magnetic field, and at that time, the current is maximum.

The total energy, when I is maximum, can be written as follows:

E = 1/2 L I² (1)

In our case, when I= 2.4A, E= 56 mJ.

So, we can find out the value of L, which will allow us to know the value of the magnetic energy at any time, having the value of the instantaneous current.

Solving for L in (1):

L = 2 *.56 mJ / (2.4)² A² = 20 mH

The next step is getting the value of the energy stored in the inductor, when I = 1.2 A, as follows:

Em = 1/2 *20 mH.* (1.2)² A² = 14.4 mJ

As the total energy must be always the same, i.e., 56 mJ, the energy stored in the capacitor, assuming no losses, must be the difference between the total energy and the one stored in the magnetic field:

Ec = 56 mJ - 14.4 mJ = 41.6 mJ

3 0
3 years ago
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