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2 years ago

Why are intensive properties useful for identifying a substance?

Chemistry

1 answer:

Lesechka [4]2 years ago

3 0

Answer:

They do not depend on the amount of substance you have.

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Charra [1.4K]

Here I hope this helps

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2 years ago

Is a push or a pull. A.FORCE B.MASS ???

Alex_Xolod [135]

A.Force because your adding pressure to what you are pushing or pulling.

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2 years ago

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

8 0

2 years ago

What elements were formed in Nuclear fusion?

Sonja [21]

Answer:

tritium and deuterium are combined and result in the formation of helium

6 0

2 years ago

Consider the following electron configurations to answer the questions that follow: (i) 1s2 2s2 2p6 3s1 (ii) 1s2 2s2 2p6 3s2 (ii

Ahat [919]

Option (i) would have the highest 2nd Ionization Energy.

Option (i) is Sodium.

Can be Written as 2, 8 , 1

For its 1st Ionization energy... It'd be extremely easy to remove that Electron cos its on the outermost shell.

Now After Removing that Electron...

Sodium's Electronic Configuration Reduces to that of Neon Which is 2, 8.

Neon has a very stable Octet.

It would take an ENORMOUS amount of energy to break its Octet stability... that is... Remove 1 electron from its Octet.

Option (i) [Sodium] has the highest 2nd Ionization Energy

6 0

2 years ago