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denis23 [38]
3 years ago
7

In the reaction at Blood Falls, iron and oxygen combine to form iron oxide, which is called rust (water is also present). The re

actants are ,
A] Oxygen and iron oxide
B] iron ad iron oxide
C] Oxygen and iron

and the product is

A] iron
B] Oxygen
C] Iron oxide.

i need answers to boh blanks please.
Physics
2 answers:
notka56 [123]3 years ago
8 0

Answer:The reactants are Oxygen and iron.

The product is Iron oxide.

Explanation:

When iron and oxygen reacts with each other produce iron oxide which reddish brown in color. This reddish brown iron oxide is often refereed to as Rust.

4Fe+3O_2\rightarrow 2Fe_2O_3(\text{reddish brown})

The Reactants are iron and oxygen.

The product is iron oxide

uranmaximum [27]3 years ago
3 0
Reactant is<span> a substance that is in a chemical </span>reaction<span>. Product is a substance that is produced by the chemical </span>reaction. A chemical change that you are familiar with isrust<span>. In this chemical </span>reaction<span>, </span>oxygen<span> and </span>iron<span>, which are the </span>reactants,combine to form<span> a product called </span>iron oxide(rust<span>)
(Mark me as brainiest, vote, and give thanks! Trying to rank up!)</span>
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Answer:

Q=3.47\times 10^{-15}\ C

Explanation:

Given that,

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To find the net charge if there are n number of extra electrons is :

Q = n × q

Q=21749\times 1.6\times 10^{-19}\ C

Q=3.47\times 10^{-15}\ C

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There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel
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The total electric potential at mid way due to 'q' is \frac{q}{4\pi\epsilon_{o}d}

The net Electric field at midway due to 'q' is 0.

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The electric potential at a distance d due to 'Q' is:

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Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}

Similar is the case with plate B:

V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}

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\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}

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Electric Field at plate A, \vec{E_{A}} at midway due to charge q:

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Similarly, for plate B:

\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

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