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3 years ago

The type of chemical bond that involves the transfer of electrons between atoms is a(n) bond.

Physics

2 answers:

GaryK [48]3 years ago

7 0

<h3>Answer;</h3>

<em>-Ionic bond</em>

<h3>Explanation;</h3>

<u>Ionic bond</u> is a type of bonds that results from transfers of electrons from one atom to another. This types of bond is formed between metals and non metals.
Metals lose electrons to form cations (positively charged ion) and non metals gain electrons forming anions, negatively charged ion.
For instance; between sodium and chlorine element; sodium donates a single electron to attain a stable configuration while chlorine gains an electron to achieve stable configuration, thus forming an ionic compound sodium chloride.

Brilliant_brown [7]3 years ago

6 0

The type of chemical bond that involves the transfer of electrons between atoms is a(n) ionic bond.

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For an object to appear red, what must happen?

Lapatulllka [165]

Answer:

<h2><em><u>C. It must reflect red light wavel</u></em></h2>

Explanation:

<em>The color of the object is reflected by the object other colors are absorbed by the object.</em>

Here <em>object appears in red color so red color is reflected by the object and other colors are absorbed.</em>

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2 years ago

Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter

alukav5142 [94]

Answer: 0.258

Explanation:

The resistance $R$ of a wire is calculated by the following formula:

$R=\rho\frac{l}{s}$ (1)

Where:

$\rho$ is the resistivity of the material the wire is made of. For aluminium is $\rho_{Al}=2.65(10)^{-8}m\Omega$ and for copper is $\rho_{Cu}=1.68(10)^{-8}m\Omega$

$l$ is the length of the wire, which in the case of aluminium is $l_{Al}=12m$ , and in the case of copper is $l_{Cu}=30m$

$s$ is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

$s=\pi{(\frac{d}{2})}^{2}$ (2) Where $d$ is the diameter of the circumference.

For aluminium wire the diameter is $d_{Al}=2.5mm=0.0025m$ and for copper is $d_{Cu}=1.6mm=0.0016m$

So, in this problem we have two transversal areas:

<u>For aluminium:</u>

$s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}$

$s_{Al}=0.000004908m^{2}$ (3)

<u>For copper:</u>

$s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}$

$s_{Cu}=0.00000201m^{2}$ (4)

Now we have to calculate the resistance for each wire:

<u>Aluminium wire:</u>

$R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}$ (5)

$R_{Al}=0.0647\Omega$ (6) Resistance of aluminium wire

<u>Copper wire:</u>

$R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}$ (6)

$R_{Cu}=0.250\Omega$ (7) Resistance of copper wire

At this point we are able to calculate the ratio of the resistance of both wires:

$Ratio=\frac{R_{Al}}{R_{Cu}}$ (8)

$\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}$ (9)

Finally:

$\frac{R_{Al}}{R_{Cu}}=0.258$ This is the ratio

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Yanka [14]

Answer:

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