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andreyandreev [35.5K]
3 years ago
11

My experiment is:

Chemistry
1 answer:
Dmitriy789 [7]3 years ago
8 0
The independent variable is the amount of salt added to the ice. The dependent variable is the amount of time it takes for the ice to completely melt.
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The most reactive metals are located at the a)top b)far-left c)far-right d)center of the periodic table.
Jlenok [28]

Answer:

Bottom left corner of the periodic table

Explanation:

The elements toward the bottom left corner of the periodic table are the metals that are the most active in the sense of being the most reactive. Lithium, sodium, and potassium all react with water, for example.

7 0
4 years ago
Read 2 more answers
Which if the following numbers in your personal life are exact numbers?
mr_godi [17]

Answer:

I think D

Explanation:

4 0
3 years ago
Which of the following is not an example of a chemical change?
Eva8 [605]
D I think.. hopefully it helps a little
8 0
3 years ago
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.

6 0
4 years ago
What is the formula of nitrous acid
Fudgin [204]

Answer:

HNO2

Explanation:

It is a pale blue solution and it is only known in solution and gas form.

4 0
3 years ago
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