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3 years ago

5

When the ideal gas law is arranged as shown below, what property of the gas is being solved for (represented by the X)? x equals

PV over nR

1 answer:

Mkey [24]3 years ago

3 0

That's kind of a ponderous way to describe it, but your 'X' represents
the absolute temperature of the ideal gas.

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Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

<u>Answer:</u> The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

<u>Explanation:</u>

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

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

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

<u>Case 1:</u>

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

<u>Case 2:</u>

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

<u>Case 3:</u>

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

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3 years ago

A fish is swimming at a constant rate towards the ocean floor. The equation

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A ﬁsh is swimming at a constant rate toward the ocean ﬂoor. The equation y = −7x − 3 can be used to represent this situation, where y is the depth of the ﬁsh in meters below sea level and x is the number of seconds the ﬁsh has been swimming. Which statement best describes the depth of the ﬁsh, given this equation?
From a starting position of 7 meters below sea level, the ﬁsh is descending 3 meters per second.
B From a starting position of 7 meters below sea level, the ﬁsh is ascending 3 meters per second.
C From a starting position of 3 meters below sea level, the ﬁsh is descending 7 meters per second

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Here we have to draw the mechanism of the reduction reaction between benzaldehyde and sodium borodeuteride to form the corresponding alcohol.

The reducing agent sodium borodeuteride can reduce the aldehydes to its corresponding alcohol. The reaction mechanism is shown in the attached image.

The reaction mechanism can be explained as-

The sodium borodeuteride is highly ionic in nature thus it remains as Na⁺ and BD₄⁻ The deuterium atom of BD₄⁻ attack the carbonyl carbon atom and substitute one of its deuterium as shown in the figure.

One molecule of sodium borodeuteride can reduce four molecules of benzaldehyde. The polar solvent like alcohol donates the proton as shown in the mechanism.

The converted alcohol contains the deuterium atom at the -C center. Thus benzaldehyde is converted to deuteroted benzyl alcohol.

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CHM 17 Practice Final Exam

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Answer:

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Explanation:

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For some jobs, “good enough” is good enough. That’s sometimes true in chemistry. More often, though, careful planning, calculati

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When carrying out experiment in the laboratory to investigate physical changes, absolute measurement is usually not needed. For instance, one can just take a little sample of sodium chloride and dissolve it in water to demonstrate physical change. But for other experiments, such as the measurement of the catalase enzyme activity in a liver tissue. Careful measurements have to be made when preparing the chemicals that will be used in the experiment.

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