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3 years ago

The periodic table shows metals, nonmetals, and noble gases. Which two elements will most likely form an ionic bond?

Physics

2 answers:

sertanlavr [38]3 years ago

5 0

<h3><u>Answer;</u></h3>

<em>B. Magnesium and silicon </em>

<h3><u>Explanation;</u></h3>

<em><u>Ionic bond is a type of bond that is formed between a metal and a non metal. </u></em>This type of bond occurs as a result of transfer of electrons where an atom of a metal looses electrons and that of a non metal gains electrons. Consequently, t<em><u>he atoms involved attain a stable configuration, where one becomes a cation and the other becomes an anion.</u></em>
In this case, <em><u>magnesium and silicon will form an ionic bond</u></em> because magnesium is a metal and silicon is a non metal, such that <em><u>magnesium requires to loose two electrons and silicon requires to gain four electrons to attain a stable configuration and form an ionic compound.</u></em>

LiRa [457]3 years ago

3 0

<h3>The correct answer from the answers provided is <u>B) Magnesium and Silicon.</u></h3>

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B. What information did Wegener not have that would have strengthened his argument? (3

VMariaS [17]

<h2>Movement of Continents</h2>

Explanation:

Alfred Wegener was a German astronomer who proposed the "continental drift" hypothesis in 1910.
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3 years ago

A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ

Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3$

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m$

(a) the fundamental frequency is calculated as;

$f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o = 65 \ Hz$

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

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2 years ago

A circuit component with a high resistance will have a low electric current.

Zepler [3.9K]

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3 years ago

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drek231 [11]

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3 years ago

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Anvisha [2.4K]

Answer:

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hope this helps u

Explanation:

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3 years ago