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Luden [163]
3 years ago
7

When a substance is oxidized it is called a(n): select one:

Chemistry
1 answer:
BlackZzzverrR [31]3 years ago
4 0
When a substance is oxidized, its oxidation number increases. The reduction in oxidation number is due to lose of electron(s) to another substance in the reactant side. Substance which gets oxidize must reduce others. Thus it is a reducing agent.
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Consider the following reaction:
Katyanochek1 [597]

Answer:

The three statements are true

Explanation:

For the reaction:

I₂O₅(s) + 5CO(g) → I₂(s) + 5CO₂(g)

State oxidation of iodine in I₂O₅ is:

5 O²⁻ = 10⁻

As you have 2 I and the molecule has no charge, <em>oxidation state of I is +5</em>.

The carbon in CO has an oxidation state of +2 and in CO₂ is +4. That means <em>the carbon is oxidized</em>

<em />

An oxidizing agent is a substance that produce the oxidation of the agent that reacts with this one. CO is oxidized because of I₂O₅ is producing its oxidation being <em>the oxidizing agent</em>

<em></em>

Thus,<em> the three statements are true</em>.

3 0
3 years ago
Volume of a cone is 200.96 what is the height of the cone
Sedaia [141]
Is it inches millimeters meters ? 

8 0
2 years ago
HELP HELP <br><br> why are more common bone fractures among senior citizens than among young people
UkoKoshka [18]
I think this because they are getting much older and weaker so they’re bones tend to break down easily
8 0
3 years ago
Read 2 more answers
Plss answer the question in the picture. If you actually answer I will give you brainilest!!!
DedPeter [7]

Answer:

hey :)

your answer is D.

Explanation:

To find density, you must divide mass and volume.

8 divided by 4 is 2

6 divided by 6 is 0

4 divided by 2 is 2

hope i helped any :))

6 0
3 years ago
What is the ph of a solution of 0.550 m k2hpo4, potassium hydrogen phosphate?
Solnce55 [7]
We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+  + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:

PH= -㏒[H+]
     = -㏒(4.81x10^-7) = 6.32

3 0
3 years ago
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