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3 years ago

Can anyone pls help!!!!!!1

Chemistry

1 answer:

Kruka [31]3 years ago

7 0

Answer:

Well I know that when the liquid changes from clear to dark blue and small bubbles form and rise to the surface that is a reaction

Explanation:

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What is the calculated value of the cell potential at 298K for an

Tpy6a [65]

The question is incomplete, here is the complete question:

What is the calculated value of the cell potential at 298 K for an electrochemical cell with the following reaction, when the H₂ pressure is 6.56 x 10⁻² atm, the H⁺ concentration is 1.39 M, and the Sn²⁺ concentration is 9.35 x 10⁻⁴ M?

$2H^+(aq)+Sn(s)\rightarrow H_2(g)+Sn^{2+}(aq)$

Answer: The cell potential of the given electrochemical cell is 0.273 V

Explanation:

For the given chemical equation:

$2H^+(aq)+Sn(s)\rightarrow H_2(g)+Sn^{2+}(aq)$

The half cell reactions for the given equation follows:

Oxidation half reaction: $Sn(s)\rightarrow Sn^{2+}(aq)+2e^-;E^o_{Sn^{2+}/Sn}=-0.14V$

Reduction half reaction: $H_2+2e^-\rightarrow H_2(g);E^o_{2H^{+}/H_2}=0.0V$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.0-(-0.14)=0.14V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Sn^{2+}]\times p_{H_2}}{[H^+]^2}$

where,

$E_{cell}$ = electrode potential of the cell = ?

$E^o_{cell}$ = standard electrode potential of the cell = +0.14 V

n = number of electrons exchanged = 2

$[H^{+}]=1.39M$

$[Sn^{2+}]=9.35\times 10^{-4}M$

$p_{H_2}=6.56\times 10^{-2}atm$

Putting values in above equation, we get:

$E_{cell}=0.14-\frac{0.059}{2}\times \log(\frac{(9.35\times 10^{-4})\times (6.56\times 10^{-2})}{(1.39)^2})\\\\E_{cell}=0.273V$

Hence, the cell potential of the given electrochemical cell is 0.273 V

4 0

3 years ago

Convert the following to number of moles: 4.03 x 10^21 molecules present in AgNO3

castortr0y [4]

Moles = 4.03 x 10^21
------------------------
6.02 x 10^23

= 6.69 x 10^-3 moles.
Hope this helps!

6 0

3 years ago

HOCl dissociative into which 2 ions

trasher [3.6K]

Answer:

H+ and OCl-

Since this is an acid, the H+ will come off and what is left OCl- is the other part.

3 0

3 years ago

he equilibrium constant, Kc , for the following reaction is 7.00×10-5 at 673 K.NH4I(s) NH3(g) + HI(g)If an equilibrium mixture o

frosja888 [35]

Answer: The number of moles of HI in the solution is $1.24\times 10{-3}$ moles.

Explanation:

We are given:

$K_c=7.00\times 10^{-5}\\n_{NH_3}=0.405mol\\n_{NH_4I}=1.45mol\\V=4.90L$

To calculate the concentration of a substance, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ ......(1)

Concentration of ammonia:

$[NH_3]=\frac{0.405mol}{4.90L}=0.083mol/L$

Concentration of ammonium iodide:

$[NH_4I]=\frac{1.45mol}{4.90L}=0.30mol/L$

For the given chemical reaction:

$NH_4I(s)\rightleftharpoons NH_3(g)+HI(g)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[HI][NH_3]}{[NH_4I]}$

Putting values in above equation, we get:

$7.0\times 10^{-5}=\frac{[HI]\times 0.083}{0.30}$

$[HI]=2.53\times 10^{-4}$

Calculating the moles of hydrogen iodide by using equation 1, we get:

$2.53\times 10^{-5}=\frac{\text{Moles of HI}}{4.9}\\\\\text{Moles of HI}=1.24\times 10^{-3}$

Hence, the number of moles of HI in the solution is $1.24\times 10{-3}$ moles.

3 0

3 years ago

Which of the following is the correct formula for potassium nitrite? kn kno kno2 k2no

Romashka [77]

Good morning!

The correct formula for potassium nitrite is KNO2.

Hugs!

4 0

4 years ago

Read 2 more answers