PRE

Please help I just need those last 4! :)

mojhsa [17]

Answer:

Series:

3. Thermistor

Parallel:

1. ???

2. ???

3. ???

5 0

3 years ago

A 520-gram sample of seawater contains 0.317 moles of NaCl. What is the percent composition of NaCl in the water?

Volgvan

Answer:

c

Explanation:

8 0

3 years ago

Test anxiety symptoms can include

AfilCa [17]

Answer:

D.) All of the above

Explanation:

Anxiety can cause sweating, chaky hands, headaches etc

4 0

3 years ago

Read 2 more answers

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

A certain gas is found in the exhaust of automobiles and power plants. Consider a 1.53 L sample of a gas at a pressure of 5.6 x

Olegator [25]

The new volume of the gas that has an initial pressure of 5.6 x 10⁵ Pa is 5.7L. Details about volume of the gas can be found below.

<h3>How to calculate volume?</h3>

The volume of a gas can be calculated using the Boyle's law equation as follows:

P1V1 = P2V2

Where;

P1 = initial pressure
P2 = final pressure
V1 = initial volume
V2 = final volume

1.53 × 5.6 × 10⁵ = 1.5 × 10⁵ × V2

8.568 × 10⁵ = 1.5 × 10⁵V2

V2 = 8.57 × 10⁵ ÷ 1.5 × 10⁵

V2 = 5.7L

Therefore, the new volume of the gas that has an initial pressure of 5.6 x 10⁵ Pa is 5.7L.

Learn more about volume at: brainly.com/question/12357202

#SPJ1

6 0

2 years ago