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ruslelena [56]
3 years ago
6

Question 5 options: The cell potential for an electrochemical cell with a Zn, Zn2 half-cell and an Al, Al3 half-cell is _____ V.

Enter your answer to the hundredths place and do not leave out a leading zero, if it is needed.
Chemistry
1 answer:
Mashcka [7]3 years ago
4 0

Answer:

The voltage is  E_{cell} =  0.944 \ V

Explanation:

Generally the half reaction for Zn, Zn2 half-cell is mathematically represented as

    Zn_{(s)}   ⇔   Zn^{2+}_{ (aq)} + 2e^-    (reference study academy)

and the electric potential for this is a constant value

     E_{zn } =  -0.7618 \ V

Generally the half reaction for Al, Al3 half-cell is mathematically represented as

      Al^{3+}  _{(aq)} + 3e^-  ⇔  Al_{(s)}

and the electric potential for this is constant value  

       E_{Al  } =  -1.662 \ V

Therefore the cell potential for an electrochemical cell  is mathematically represented as

          E_{cell} =  E_{zn }  -  E_{Al }

substituting values

         E_{cell} =  -0.718  - (-1.662)

        E_{cell} =  0.944 \ V

 

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A 3.00-L flask is filled with gaseous ammonia, NH3. The gas pressure measured at 27.0 ∘C is 2.55 atm . Assuming ideal gas behavi
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Answer : The mass of ammonia present in the flask in three significant figures are, 5.28 grams.

Solution :

Using ideal gas equation,

PV=nRT\\\\PV=\frac{w}{M}\times RT

where,

n = number of moles of gas

w = mass of ammonia gas  = ?

P = pressure of the ammonia gas = 2.55 atm

T = temperature of the ammonia gas = 27^oC=273+27=300K

M = molar mass of ammonia gas = 17 g/mole

R = gas constant = 0.0821 L.atm/mole.K

V = volume of ammonia gas = 3.00 L

Now put all the given values in the above equation, we get the mass of ammonia gas.

(2.55atm)\times (3.00L)=\frac{w}{17g/mole}\times (0.0821L.atm/mole.K)\times (300K)

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Therefore, the mass of ammonia present in the flask in three significant figures are, 5.28 grams.

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3 years ago
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A. 1.0 liter of a 1.0 M mercury (II) chloride (HgCl2) solution<br> IMO)
insens350 [35]

What mass of the following chemicals is needed to make the solutions indicated?

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271.6g

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Given parameters:

Volume of solution  = 1L

Molarity of HgCl₂  = 1M

    number of moles of HgCl₂  = molarity of solution x volume

                                                   =   1 x 1

                                                    = 1 mole

From;

           Mass of a substance  = number of moles x molar mass;

  we can find mass;

          Molar mass of HgCl₂  = 200.6 + 2(35.5)  = 271.6g/mol

       Mass of the substance  = 1 x 271.6  = 271.6g

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