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3 years ago

Rifampin has a long aliphatic chain forming a bridge between an aromatic moiety (naphthaline).

Chemistry

1 answer:

lesya692 [45]3 years ago

7 0

Answer:

True

Explanation:

Question - Rifampin has a long aliphatic chain forming a bridge between an aromatic moiety (naphthaline).

Solution -

It is a true statement.

So,

The correct option is - True.

Reason -

Rifampin has a long aliphatic chain forming a bridge between an aromatic moiety (naphthaline)

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For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia

Fittoniya [83]

Answer : The value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

To calculate the $E^o_{(Cu^{2+}/Cu)}$ of the reaction, we use the equation:

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

$E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}$

Putting values in above equation, we get:

$1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)$

$E^o_{(Cu^{2+}/Cu)}=0.34V$

Hence, the value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

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2 years ago

MgX2 Very low density Highly reactive Diatomic Using the periodic table, determine which element these characteristics MOST LIKE

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Answer:

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Explanation:

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