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3 years ago

Is berkelium flammable

Chemistry

2 answers:

dusya [7]3 years ago

8 0

Yes berkelium is highly flammable

Brrunno [24]3 years ago

4 0

Yes. Berkelium is flammable

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What is the voltage of the voltaic cell Zn|Zn2+||Cu2+|Cu at 298 K if [Zn2+] = 0.2 M and [Cu2+] = 4.0 M? Cu2+ + 2e- → Cu Eo = +0.

OLga [1]

Answer : The voltage of the voltaic cell is 1.14 V

Explanation :

From the given cell representation, we conclude that

The 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 $E^o{cell}$ 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:

$E^o_{cell}=(+0.34V)-(-0.76V)$

$E^o_{cell}=1.1V$

Now we have to calculate the emf or voltage of the cell.

Using Nernest equation at 298 K :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Cu^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = ?

Now put all the given values in the above equation, we get:

$E_{cell}=1.1-\frac{0.0592}{2}\log \frac{0.2}{4.0}$

$E_{cell}=1.14V$

Therefore, the voltage of the voltaic cell is 1.14 V

5 0

3 years ago

The atomic radius of main group elements generally increases down a group because

Pachacha [2.7K]

The atomic radius of main group elements generally increases down a group because as there are more electrons they are farther away from the nucleus and the electrons closer to the nucleus shield the outer electrons from the protons for attraction.

8 0

3 years ago

Calculate the number of atoms in 2 mol of sulfur dioxide (SO2).

Alex17521 [72]

Two moles of SO2 means 1.204x10^24 molecules, since there are 3 atoms in one molecule, multiply 1.204 x10^24 by 3 and you get 3.612x10^24.

7 0

3 years ago

4.50 g of NaCl reacts with 10.00 g of AgNO3 to produce 7.93 g of AgCl. If the theoretical amount of AgCl that can be formed is 8

olga nikolaevna [1]

Answer: 94.07%

Explanation:

Percentage yield can be calculated by the formula

%yield = Experimental yield/Theoretical yield x100

Experimental yield = 7.93g

Theoretical yield = 8.43

%yield = Experimental yield/Theoretical yield x100

%yield = 7.93/8.43 x 100 = 94.07%

4 0

3 years ago

According to early chemists, which substances were classified as elements?

ladessa [460]

The best answer to your question would be B:those who could not be broken down by physical means.

6 0

3 years ago