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2 years ago

13

A 10.0 mL sample of 5.00M HCl is diluted until its final concentration is 0.215M. What is the final volume of the diluted soluti

on?

1 answer:

uysha [10]2 years ago

5 0

The final volume of the diluted solution that initially has a volume 10.0 mL sample and concentration of 5.00M HCl is 232.6mL.

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

The volume of a solution can be calculated using the following formula:

C1V1 = C2V2

Where;

C1 = initial concentration
C2 = final concentration
V1 = initial volume
V2 = final volume

10 × 5 = 0.215 × V2

50 = 0.215V2

V2 = 50/0.215

V2 = 232.6mL

Therefore, the final volume of the diluted solution that initially has a volume 10.0 mL sample and concentration of 5.00M HCl is 232.6mL.

Learn more about volume at: brainly.com/question/1578538

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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, $6.093\times 10^{32}M$

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3 years ago

Very important !!!!!!!!!!!!

Mademuasel [1]

They speed up reactions by lowering activation energy. Many enzymes change shape when substrates bind.

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3 years ago

How many miles of lead are equal to 9.51 x 10^3 g Pb?

Lemur [1.5K]

Molar mass Pb = 207.2 g/mol

1 mole Pb ------------- 207.2
? mole Pb ------------ 9.51 x 10³

moles = 9.51 x 10³ * 1 / 207.2

moles = 9.51 x 10³ / 207.2

= 45.89 moles

hope this helps!

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3 years ago

Does neon have a larger atomic radius than argon

Lera25 [3.4K]

Answer:

No

Explanation:

Argon has a larger atomic radius

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2 years ago

An Argon laser gives off pulses of green light (wavelength = 514 nm). If a single pulse from the laser has a total energy of 10.

svetoff [14.1K]

Answer:

$n=2.59\times 10^{16}$ photons

Explanation:

$E=n\times \frac{h\times c}{\lambda}$

Where,

n is the number of photons

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

$\lambda$ is the wavelength of the light

Given that, wavelength = 514 nm = $514\times 10^{-9}\ m$

Energy = 10.0 mJ = 0.01 J ( 1 mJ = 0.001 J )

Applying the values as:-

$0.01=n\times \frac{6.626\times 10^{-34}\times 3\times 10^8}{514\times 10^{-9}}$

$\frac{19.878n}{10^{17}\times \:514}=0.01$

$n=2.59\times 10^{16}$ photons

7 0

3 years ago