What is the number of elements in 2Ca(OH)2

Write three complete and balanced pairs of below mention electrochemical half reactions. For each pair of reactions, identify wh

kirza4 [7]

Answer: a) $Cu\rightarrow Cu^{2+}+2e^-$ : anode

b. $2H^++2e^-\rightarrow H_2$ : cathode

c. $O^{2-}\rightarrow \frac{1}{2}O_2+2e^-$ : anode

Explanation:

Electrochemical cell is a device which converts chemical energy into electrical energy. It consist of two electrodes, anode and cathode.

Oxidation i.e. loss of electrons , which results in an increase in oxidation number occurs over anode.

Reduction i.e. gain of electrons, which results in decrease in oxidation number occurs over cathode.

a. $Cu\rightarrow Cu^{2+}+2e^-$ : oxidation : anode

b. $2H^++2e^-\rightarrow H_2$ : reduction : cathode

c. $O^{2-}\rightarrow \frac{1}{2}O_2+2e^-$ : oxidation : anode

6 0

3 years ago

Into what kinds of energy does a toaster convert electrical energy?

KengaRu [80]

the answer is C. light and heat

6 0

2 years ago

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Consider the balanced equation below. 4nh3 3o2 right arrow. 2n2 6h2o what is the mole ratio of nh3 to n2? 2:4 4:2 4:4 7:2

KIM [24]

The required mole ratio of NH₃ to N₂ in the given chemical reaction is 2:4.

<h3>What is the stoichiometry?</h3>

Stoichiometry of the reaction gives idea about the number of entities present on the reaction before and after the reaction.

Given chemical reaction is:

4NH₃ + 3O₂ → 2N₂ + 6H₂O

From the stoichiometry of the reaction it is clear that:

4 moles of NH₃ = produces 2 moles of N₂

Mole ratio NH₃ to N₂ is 2:4.

Hence required mole ratio is 2:4.

To know more about mole ratio, visit the below link:
brainly.com/question/504601

3 0

2 years ago

Chromium(III) oxide can be prepared by heating chromium(IV) oxide in vacuo at high temperature: 4Cr02 —2Cr2O3 +02 The reaction o

kkurt [141]

<u>Answer:</u> The theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of $CrO_2$ = 480.1 g

Molar mass of $CrO_2$ = 84 g/mol

Putting values in equation 1, we get:

$\text{Moles of }CrO_2=\frac{480.1g}{84g/mol}=5.72mol$

For the given chemical equation:

$4CrO_2\rightarrow 2Cr_2O_3+O_2$

By Stoichiometry of the reaction:

4 moles of $CrO_2$ produces 2 moles of chromium (III) oxide

So, 5.72 moles of $CrO_2$ will produce = $\frac{2}{4}\times 5.72=2.86mol$ of chromium (III) oxide

Now, calculating the mass of chromium (III) oxide from equation 1, we get:

Molar mass of chromium (III) oxide = 152 g/mol

Moles of chromium (III) oxide = 2.86 moles

Putting values in equation 1, we get:

$2.86mol=\frac{\text{Mass of chromium (III) oxide}}{152g/mol}\\\\\text{Mass of chromium (III) oxide}=(2.86mol\times 152g/mol)=434.72g$

To calculate the percentage yield of chromium (III) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of chromium (III) oxide = 402.4 g

Theoretical yield of chromium (III) oxide = 434.72 g

Putting values in above equation, we get:

$\%\text{ yield of chromium (III) oxide}=\frac{402.4g}{434.72g}\times 100\\\\\% \text{yield of chromium (III) oxide}=\%$

Hence, the theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

7 0

2 years ago

How does electrolysis work

liberstina [14]

Eeletrolysis is a method of removing individual hairs from the face or body .Medical electrolysis devices destroy the growth center of the hair with chemical or heat energy.After a very fine probe is inserted into the hair follicle, the hair is removed with tweezers

5 0

2 years ago

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