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2 years ago

When iron pipes come in contact with water and oxygen, iron transfers electrons to the oxygen molecules. Which

1 answer:

n200080 [17]2 years ago

8 0

Considering the definition of redox reactions, the correct answer is the last option: Oxygen is reduced, and iron is oxidized.

<h3>Oxidation and reduction</h3>

Oxidation is a reaction where an atom, ion or molecule loses electrons while reduction corresponds to the gain of electrons from an atom, ion or molecule.

<h3>Redox reactions</h3>

The oxidation and reduction reactions always occur simultaneously for what are generally known as oxidation-reduction reactions or redox reactions.

That is, a redox reaction is a chemical reaction that occurs between an oxidizing substance and a reducing substance. During the reaction, the oxidizing substance loses electrons and the reducing substance gains electrons.

<h3>Oxidizing and reducing agent</h3>

That is, the oxidizing agent is the one that traps the electrons while the reducing agent is the one that provides the electrons.

In this case, when iron pipes come in contact with water and oxygen, iron transfers electrons to the oxygen molecules. So, iron loses electrons while oxygen gains electrons. So oxygen is reduced, and iron is oxidized.

Then, the correct answer is the last option: Oxygen is reduced, and iron is oxidized.

Learn more about oxidation and reduction:

<u>brainly.com/question/10507355?referrer=searchResults</u>

<u>brainly.com/question/14216150?referrer=searchResults</u>

<u>brainly.com/question/10547418?referrer=searchResults</u>

<u>brainly.com/question/14096111?referrer=searchResults</u>

<u>brainly.com/question/24667543?referrer=searchResults</u>

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b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul

Nat2105 [25]

Answer:

b) Betelgeuse would be $\approx 1.43 \cdot 10^{6}$ times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is $\approx 1.37 \cdot 10^{-5} }$ the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

$B = \displaystyle{\frac{L}{4\pi d^2}}$ (1)

where $B$ is the brightness, $L$ is the star luminosity and $d$ , the distance from the star to the point where the brightness is calculated (measured). Thus:

b) $B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}}$ and $B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}}$ where $L_{Sun}$ is the Sun luminosity ( $3.9 x 10^{26} W$ ) but we don't need to know this value for solving the problem. $ly$ is light years.

Finding the ratio between the two brightness we get:

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }$

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is $1.581 \cdot 10^{-5}\ ly$ . Then

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }$

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

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3 years ago

light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance li

lapo4ka [179]

Answer:

The distance traveled in 1 year is: $3.143*10^{16}ft$

Explanation:

Given

$s = 982,080,000 ft/s$ --- speed

$t = 32,000,000 s$ --- time

Required

The distance traveled

This is calculated as:

$Speed = \frac{Distance}{Time}$

So, we have:

$Distance = Speed * Time$

This gives:

$Distance = 982,080,000 ft/s * 32,000,000 s$

$Distance = 982,080,000 * 32,000,000ft$

$Distance = 3.143*10^{16}ft$ -- approximated

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2 years ago

Starting from rest, a 2.1x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts

nirvana33 [79]

Answer:

1.327363 m/s

0.00090243 m

Explanation:

u = Initial velocity

v = Final velocity

m = Mass of flea

Energy

$E=\frac{1}{2}m(v^2-u^2)\\\Rightarrow 3.7\times 10^{-4}=2.1\times 10^{-4}(v^2-0)\\\Rightarrow v=\sqrt{\frac{3.7\times 10^{-4}}{2.1\times 10^{-4}}}\\\Rightarrow v=1.32736\ m/s$

The velocity of the flea when leaving the ground is 1.327363 m/s

$W=F\times s\\\Rightarrow s=\frac{W}{F}\\\Rightarrow s=\frac{3.7\times 10^{-4}}{0.41}\\\Rightarrow s=0.00090243\ m$

The flea will travel 0.00090243 m upward

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3 years ago

Does a basketball, baseball, tennis ball, or marble MOST LIKELY have the smallest volume?

Fantom [35]

Marble i think not quite sure doe

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3 years ago

Read 2 more answers

Consider five atoms from the second period : lithium , beryllium , boron , carbon , and nitrogen , which of these elements has t

kap26 [50]

Answer:

Lithium has the lowest electronegativity.

Explanation:

Electronegativity measures the tendency of an atom to attract the bonding pair of electrons. As we move left to right in a period, the number of shell remains same but the number of electron increases(negative charge increases. Hence, electronegativity also increases.

As Lithium is the left most element in this period. It has the lowest electronegativity value which is equal to 0.98.

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3 years ago