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djyliett [7]
4 years ago
9

Elaborate on what can be learned about the properties by the location of element on the periodic table. Use atomic number 13 as

an example. A) It is a non-metal, has 3 valence electrons, becomes a cation with a -3 charge, and is brittle, moldable, and non-conductive. B) It is a metalloid, has 3 valence electrons, becomes a cation with a -5 charge, and is workable, inelastic, and non-conductive. C) It is a main group metal, has 3 valence electrons, becomes a cation with a +3 charge, and is ductile, malleable, and conductive. D) It is a transition metal, has 1 valence electrons, becomes a cation with a +1 charge, and is ductile, malleable, and conductive. Eliminate
Physics
2 answers:
svet-max [94.6K]4 years ago
7 0

Answer:

C. It is a main group metal, has 3 valence electrons, becomes a cation with a +3 charge, and is ductile, malleable, and conductive.

Explanation:It is a main group metal, has 3 valence electrons, becomes a cation with a +3 charge, and is ductile, malleable, and conductive.

The location of element on the periodic table gives the electron configuration. From this you know the number of valence electrons, type of ion formed (cation or anion), and the charge of the ion. Additionally the location gives info about the properties because you know if the element is a metal, nonmetal, or metalloid. Using atomic number 13 as an example - Aluminum is a main group metal, has 3 valence electrons, becomes a cation with a +3 charge, and is ductile, malleable, and conductive.

aleksklad [387]4 years ago
4 0

The right answer is c

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Suppose that it takes a simple pendulum 1.2 seconds to swing from its leftmost point to its rightmost point. what is the period
Mariana [72]

The time period of the simple pendulum is 2.4 seconds.

Given the data in the question;

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  • Period of the pendulum; T = \ ?

<h3>What is Period?</h3>

Period is the time needed for a complete cycle of vibration to pass a given point.

Period of a pendulum is the of time needed for it to complete one full back-and-forth motion. It is the time required to for the pendulum to swing from leftmost point to rightmost point and back to leftmost point.

Now, if it took the pendulum 1.2s to swing from leftmost point to rightmost point, it will also take the pendulum 1.2s to swing back to its original position( leftmost point )

Hence,

T = time taken to swing from leftmost to rightmost + time taken to swing from rightmost point to leftmost point.

T = 1.2s + 1.2s\\\\T = 2.4s

Therefore, the time period of the simple pendulum is 2.4 seconds.

Learn more about Time Period: brainly.com/question/27135322

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3 years ago
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Why does the satellite not fall while revolving the earth​
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3 years ago
Determine the number of atoms per unit cell in a (a) face-centered cubic, (b) body- centered cubic, and (c) diamond lattice.
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Answer:

a) 4

b) 2

c) 8

Explanation:

In a cubic lattice, each atom of the vertex is shared among other 8 unit cells, so the atoms on the vertex contribute to 1/8 to a given unit cell.

a) Ina face-centered cubic we have 8 atoms on the vertex and one in each face, which is shared with another unit cell, so it contribute to 1/2

Therefore, the total atoms are:

8*(1/8) + 6*(1/2) = 4

b) In the body centered cubic structure, the centered atom is not shared with another cell, therefore it contribute to 1 to the given cell:

The number of atoms per unit cel is:

8*(1/8) + 1 = 2

c) The diamond lattice is similar to the face-centered cubic lattice but it contains two identical atoms per lattice point.

Therefore it must contain twice atoms than the face-centered cubic lattice:

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3 0
4 years ago
When mass m is tied to the bottom of a long, thin wire suspended from the ceiling, the wire's second-harmonic frequency is 180 h
Katena32 [7]
The frequency of the nth-harmonic of a string is given by
f_n =  \frac{n}{2L}  \sqrt{ \frac{T}{\mu} }
where n is the number of the harmonic, L is the length of the string, T the tension and \mu the linear density. 

In our problem, since the mass m is tied to the string, the tension is equal to the weight of the object tied:
T=mg
Substituting into the first formula, we have
f_n =  \frac{n}{2L}  \sqrt{ \frac{mg}{\mu} }

In our problem we have n=2 (second harmonic). In the previous equation, the only factor which is not constant between the first and the second part of the problem is m, the mass. So, we can rewrite everything as
f_2 = K  \sqrt{m}
where we called 
K= \frac{2}{2L}  \sqrt{ \frac{g}{\mu} }

In the first part of the problem, the mass of the object is m and f_2 = 180 Hz. So we can write 
180 Hz = K  \sqrt{m}

When the mass is increased with an additional 1.2 kg, the relationship becomes
270 Hz = K \sqrt{(m+1.2 Kg)}

By writing K in terms of m in the first equation, and subsituting into the second one, we get
180 Hz  \sqrt{ \frac{m+1.2 Kg}{m} }=270 Hz
and solving this, we find
m=0.96 kg


5 0
4 years ago
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