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3 years ago

PLEASE HELP ILL GIVE U BRAINLIEST What's something you learned about the periodic table

Chemistry

2 answers:

valina [46]3 years ago

7 0

Answer:

The periodic table of elements puts all the known elements into groups with similar properties. This makes it an important tool for chemists, nanotechnologists, and other scientists. If you get to understand the periodic table and learn to use it, you'll be able to predict how chemicals will behave.

alexandr1967 [171]3 years ago

4 0

Answer:

Okayy

Explanation:

The periodic table arranges the chemical elements

All the members of a family of elements have the same number of valence electrons and similar chemical properties. The horizontal rows on the periodic table are called periods

It have many properties. Each column is called a group.

ALSO Metals are placed on the left-hand side of the periodic table , and non-metals on the right.

Metalloids can also be called semimetals. On the periodic table, the elements colored yellow, which generally border the stair-step line, are considered to be metalloids.

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3 years ago

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A voltaic cell made of a Cr electrode in a solution of 1.0 M in Cr3+ and a gold electrode in a solution that is 1.0 M in Au3+.

Vikki [24]

Answer: a) Anode: $Cr\rightarrow Cr^{3+}+3e^-$

Cathode: $Au{3+}+3e^-\rightarrow Au$

b) Anode : Cr

Cathode : Au

c) $Au^{3+}+Cr\rightarrow Au+Cr^{3+}$

d) $E_{cell}=2.14V$

Explanation: -

a) The element Cr with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element Au with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.

At cathode: $Au{3+}+3e^-\rightarrow Au$

At anode: $Cr\rightarrow Cr^{3+}+3e^-$

b) At cathode which is a positive terminal, reduction occurs which is gain of electrons.

At anode which is a negative terminal, oxidation occurs which is loss of electrons.

Gold acts as cathode ad Chromium acts as anode.

c) Overall balanced equation:

At cathode: $Au{3+}+3e^-\rightarrow Au$ (1)

At anode: $Cr\rightarrow Cr^{3+}+3e^-$ (2)

Adding (1) and (2)

$Au^{3+}+Cr\rightarrow Au+Cr^{3+}$

d) $E^0_(Cr^{3+}/Cr)$ = -0.74 V

$E^0_(Au^{3+}/Au)$ = 1.40 V

$E^0{cell}=E^0{cathode}-E^0{anode}=1.40-(-0.74)=2.14V$

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Au^{3+}]}{[Cr^{3+}]^}$

where,

n = number of electrons in oxidation-reduction reaction = 3

$E^o_{cell}$ = standard electrode potential = 2.14 V

$E_{cell}=2.14-\frac{0.0592}{3}\log \frac{[1.0}{[1.0]}$

$E_{cell}=2.14$

Thus the standard potential for an electrochemical cell with the cell reaction is 2.14 V.

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4 years ago

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blagie [28]

Answer:

the correct answer is E

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