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4 years ago

Which of the following describes a synthesis reaction?

Chemistry

1 answer:

Grace [21]4 years ago

7 0

Answer:

D. Two reactants combine to form one product

Explanation:

The definition of synthesis in chemistry is "the production of chemical compounds by reaction from simpler materials."

A is decomposition, B is DOUBLE replacement, and C is SINGLE replacement.

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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 :

<u>(a) As per question, lead is oxidized and copper is reduced.</u>

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

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

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

Which part of the atom cannot have its location accurately determined and is modeled by a cloud around the center of the atom? A

Tju [1.3M]

Answer:

Electron

Explanation:

The answer would be the electron because it is constantly moving so its location cannot be accurately determined

3 0

3 years ago

Read 2 more answers

What volume of 0.160 m li2s solution is required to completely react with 130 ml of 0.160 m co no3 2?

Over [174]

The volume of 0.160 m Li2S solution required to completely react with 130 ml of 0.160 CO(NO3)2 is calculated as below

write the reacting equation

Co(NO3)2 + Li2S = 2LiNO3 + COS

find the moles of CO(NO3)2 = molarity x volume

= 130 ml x 0.160=20.8 moles

since the reacting moles between CO(NO3)2 to LiS is 1:1 the moles of LiS is also 20.8 moles

volume of Lis is therefore = moles of Lis/ molarity of LiS

= 20.8/0.160 = 130 Ml

3 0

3 years ago

A solution of rubbing alcohol is 68.6 % (v/v) isopropanol in water. How many milliliters of isopropanol are in a 88.2 mL sample

KonstantinChe [14]

From the information given, the total volume of rubbing alcohol is 88.2 ml

68.6 % of this volume is isopropanol.

We will assume 88.2 ml represents 100% volume, so the volume of water will be 31.4 %

The volume of isopropanol is

68.6/100 x 88.2 → 0.686 × 88.2 = 60.505 ml

The volume of isopropanol is 60.5 ml.

Volume of water will be 88.20 - 60.5 = 27.7 ml

(27.7 / 88.2 × 100 = 31.4% )

Adding 60.5 ml of isopropanol to 27.7 ml of water to make up 88.2 ml will give 68.6 % v/v isopropanol to water solution.

8 0

4 years ago

When concentrated nitric acid mixed with some concentrated sulfuric acid reacts with benzene at moderate temperature forms------

FrozenT [24]

I will help just send me it

5 0

3 years ago