All categories

3 years ago

I NEED HELP PLEASE, THANKS! :)

Chemistry

1 answer:

Free_Kalibri [48]3 years ago

7 0

Answer:

Explanation:

Metallic character increases across a period to the left and downwards.

If you look at the periodic table, Cs is lower and more towards the left.

You might be interested in

Wsp im fisdes once again hiiiii

Diano4ka-milaya [45]

Answer:

heyyyyy

Explanation:

6 0

3 years ago

Read 2 more answers

Which of the following is a unit for volume? m3 km cg dkm

babymother [125]

M³ is a unit for volume.

8 0

3 years ago

Read 2 more answers

True or False- The parts of a mixture keep their own properties when they are combined.

ICE Princess25 [194]

It is true yes :) happy to help

4 0

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

Convert 1.580 hectoliters to deciliters. liters

7nadin3 [17]

Answer:

1580 decilliters

Explanation:

5 0

2 years ago

Read 2 more answers