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2 years ago

14

Una masa de hidrogeno H2 ocupa un volumen de 12 litros a 730 mmHg. ¿Cuál es el volumen del gas a 1200 mmHg, si la temperatura pe

rmanece constante?

1 answer:

Alex2 years ago

4 0

Answer:

ENGLISH PLEASE

Explanation:

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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

Fatima is designing an advanced graphics card for a next-generation video game system. She needs to be able to

Bad White [126]

Answer:

semiconducting,tellurium

Explanation:

just completed the assignment

4 0

3 years ago

Read 2 more answers

Zinc oxide adopts a face-centered cubic arrangement. Both Zinc ions and oxide ions adopt the face-centered cubic arrangement; wi

vichka [17]

Answer:

5.41 ×10⁻²²

Explanation:

We were told right from the question that both the Zinc ions and the Zinc oxide adopts a face-centered cubic arrangement.

Then, the number of ZnO molecule in one unit cell = 4

The standard molar mass of ZnO = 81.38g

Avogadro's constant = 6.023 × 10²³ mole

∴

The mass of one unit cell of zinc oxide can be calculated as:

= $\frac{4*81.38}{6.023*10^{23}}$

= 5.40461564×10⁻²²

≅ 5.41 ×10⁻²²

∴ The mass of one unit cell of zinc oxide = 5.41 ×10⁻²²

3 0

2 years ago

What is the mass percentage for copper ?

RideAnS [48]

Answer:

Explanation:

To find the mass percent composition of an element, divide the mass contribution of the element by the total molecular mass. This number must then be multiplied by 100% to be expressed as a percent.

3 0

2 years ago

19. What is sthe mass of 6.02 x 1024 atoms

alekssr [168]

The mass of magnesium in $6.02 \times 10^{24}$ atoms is 240 g.

Answer: Option A

<u>Explanation:</u>

First, we have to convert the atoms to moles of magnesium.

We know that $6.02 \times 10^{24}$ atoms are present in 1 mole of magnesium. So,

$\text { 1 atom }=\frac{1}{6.02 \times 10^{24}} \text { moles }$

$6.02 \times 10^{24} \text { atoms }=\frac{6.02 \times 10^{24}}{6.02 \times 10^{23}} \text { moles }=10 \text { moles }$

Thus,

$Mass of M g= molar mass of M g \times No. of moles of M g$

$\text { Mass of } M g=24 \times 10=240 g$

Thus, 240 g of Magnesium is present in $6.02 \times 10^{24}$ atoms.

4 0

3 years ago