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3 years ago

Describe a situation where scientific knowledge was gained through doing an experiment

2 answers:

8 0

Scientific knowledge can be gained through experiment by doing a thorough experimental procedure that will enable to answer the objectives of the experiment that will lead to a scientific knowledge. Through experimentation you can know the behavior of a variable with respect to a certain variable. This can also answer some knowledge gap about certain topic

ser-zykov [4K]3 years ago

8 0

Answer:

When determining the types of chemical reactions.

Explanation:

Hello,

In this case, in order to establish the different types of chemical reactions, several experiments are carried out the determine whether a chemical reaction is synthesis, decomposition, double displacement, double displacement or combustion. Such experiments consist on the combination of different of reactants under varied conditions in order the appreciate the reaction behavior over the time, changes in color, temperature changes and formed products that suggest the type of reaction.

Once the observations are set throughout the experiments, one provides the scientific knowledge that in this case will help to differentiate such types of chemical reactions.

Best regards.

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Answer: The correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

Explanation:

We are given:

$Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: $Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V$

Reduction half reaction: $Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V$ ( × 3)

Net equation: $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.80-(-0.73)=1.53V$

Hence, the correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

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