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3 years ago

To create hydrogen and oxygen from water, you would need to perform which type of reaction?

Chemistry

2 answers:

Murrr4er [49]3 years ago

6 0

B. Decomposition is the correct answer~

Lana71 [14]3 years ago

6 0

Answer is: B. Decomposition.

Decomposition is reaction where one substance is broken down into two or more simpler substances.

Electrolysis of water is the decomposition reaction, because from one molecule (water) two molecules (hydrogen and oxygen) are produced.

Water is separeted into two molecules:

Reaction of reduction at cathode: 2H⁺(aq) + 2e⁻ → H₂(g).

Reaction of oxidation at anode: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻.

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Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Part (a) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

Part (b) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

5 0

3 years ago

The fifth element in group 7 in astatine would you expect it to be gas, liquid, or solid? why?

Ymorist [56]

Answer:

solid

Explanation:

Melting and boiling points of Group 7 elements State at room temperature Room temperature is usually taken as being 25°C. At this temperature, fluorine and chlorine are gases, bromine is a liquid, and iodine and astatine are solids. There is therefore a trend in state from gas to liquid to solid as you go down the group.

6 0

2 years ago

True or false: the most easily ionizable elements are the most electronegative.

ehidna [41]

False, This is because when you can easily ionize and atom or the chances of it being ionizable are quite high, it means that that particular atom have very low ionization potential that is the reason why it was easily ionizable An atom with a high ionization power and a firmly negative electron fondness will both pull in electrons from different particles and oppose having its electrons taken away; it will be an exceedingly electronegative molecule.

3 0

3 years ago

When temperature drops, (for example from 20 degrees celsius to 10 degrees celsius)

AnnZ [28]

The answer should be B !

7 0

3 years ago

Read 2 more answers

Which of the following interactions can contribute to the intrinsic binding energy during enzymatic catalysis? choose all that a

Lady_Fox [76]

1. electrostatic interactions
3. van de waals interactions 
4. hydrogen bonding

7 0

3 years ago

Read 2 more answers