Carbon tetrachloride is made according to the following reaction:

Which statement is correct about the rate of chemical reactions?

ratelena [41]

Answer:

It increases when a catalyst is added.

Explanation:

The following factors control reaction rates:

1. Nature of reactants

2. Concentration of the reactants or pressure of gaseous

3. Temperature

4. Presence of catalyst

5. Sunlight

The addition of a foreign body to a reaction may influence the speed of the reaction. If a foreign body increases the rate of reaction, it is a called a positive catalyst or simply a catalyst. A negative catalyst is called an inhibitor.

A catalyst is a substance that is introduced into a chemical reaction to change the rate of the reaction without itself being affected at the end of the reaction.

Catalysts helps to reduce reaction time of many slow reactions. Most catalysts are specific in their actions and works on certain reactions or substrates.

Temperature change has a considerable effect on reaction rates since temperature is directly proportional to the average kinetic energy of reacting particles. Generally, reaction rate varies as temperature directly.

8 0

3 years ago

An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6

ExtremeBDS [4]

Answer :

(a) The concentration of $Pb^{2+}$ is, 0.0337 M

(b) The concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

Solution :

(a) As per question, lead is oxidized and copper is reduced.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Pb\rightarrow Pb^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

The balanced cell reaction will be,

$Pb(s)+Cu^{2+}(aq)\rightarrow Pb^{2+}(aq)+Cu(s)$

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}$

$E^o=0.34V-(-0.13V)=0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

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

where,

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

$E_{cell}$ = 0.507 V

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

$0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}$

$[Pb^{2+}]=0.0337M$

Therefore, the concentration of $Pb^{2+}$ is, 0.0337 M

(b) As per question, lead is reduced and copper is oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction: $Pb^{2+}+2e^-\rightarrow Pb$

The balanced cell reaction will be,

$Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)$

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=-0.13V-(0.34V)=-0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

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

where,

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

$E_{cell}$ = 0.507 V

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

$0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}$

$[Pb^{2+}]=6.093\times 10^{32}M$

Therefore, the concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

6 0

3 years ago

It is necessary to have a 40% antifreeze solution in the radiator of a certain car. the radiator now has 70 liters of 20% soluti

Akimi4 [234]

The answer is : 17.5 liters drained and replaced by 17.5 liters of 100% solution.
x = amount drained and replaced
(70-x) = remaining amount of 20% solution
.20(70-x) + 1.00(x) = .40(70)
14 - .2x + 1x = 28
1x - .2x = 28 - 14
.8x = 14
x = 14/.8
x= 17,5 ( 17.5 liters drained and replaced by 17.5 liters of 100% solution)

6 0

3 years ago

Read 2 more answers

`You have to be careful about pouring drano down your pipes since it is mainly hydrochloric acid--you can't do it if they are ma

Zanzabum

Answer:

6.67 moles

Explanation:

Given that:-

Moles of hydrogen gas produced = 10.0 moles

According the reaction shown below:-

$2Al + 6HCl\rightarrow 2AlCl_3 +3H_2$