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3 years ago

Balance the equation 2kclo3 3o2 + kcl

Chemistry

1 answer:

sergij07 [2.7K]3 years ago

5 0

Answer is 2KClO3 3O2 + 2KCl

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34 grams of carbon react with an unlimited amount of H2O. The reaction is: C + H2O → CO + H2 What is the ending substance?

katovenus [111]

H2c+o
hope this helps :0

7 0

3 years ago

Read 2 more answers

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

3 years ago

When lithium reacts with bromine to form the compound LiBr, each lithium atom:

yKpoI14uk [10]

(3) loses one electron and becomes positively charged
Lithium has one valence electron and Bromine has seven. Therefore Lithium will give up its one to Bromine for both to have an octet

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3 years ago

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The __________________ of a solution tells how many moles of solute are present per liter.

Stels [109]

Answer:

The MOLARITY of a solution tells how many moles of solute are present per liter

Explanation:

Molarity is a sort of concentration.

It is written as M (mol/L)

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3 years ago