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SashulF [63]
2 years ago
5

A voltaic cell consists of an Mn/Mn2 half-cell and a Pb/Pb2 half-cell. Calculate [Pb2 ] when [Mn2 ] is 1.1 M and E cell is 0.44

V.
Chemistry
1 answer:
djyliett [7]2 years ago
8 0

Answer:

[Pb²⁺] = 2.31x10⁻²¹ M

Explanation:

Let's write the semi reaction for each half cell:

Pb²⁺ + 2e⁻ -----------> Pb(s)     E° = -0.13 V

Mn²⁺ + 2e⁻ ----------> Mn(s)     E° = -1.18 V

As we can see, the E° of Pb is higher than the E° of the Mn, thus, Pb is reducting and Mn is oxidizing:

Pb²⁺ + 2e⁻ -----------> Pb(s)     E°₁ = -0.13 V

Mn(s) ---------> Mn²⁺ + 2e⁻      E°₂ = +1.18 V

E° = E°₁ + E°₂

E° = -0.13 + 1.18 = 1.05 V

Now, we can use the Nerst equation which is:

E = E° - 0.059/n log([Mn²⁺] / [[Pb²⁺])

From here, we just need to replace and then, solve for the [Pb²⁺]:

0.44 = 1.05 - 0.059/2 log(1.1 / x)

0.44 - 1.05 = -0.0295 log(1.1 / x)

-0.61 / -0.0295 = log(1.1 / x)

antlog(20.678) = 1.1 /x

x = [Pb²⁺] = 1.1 / 4.76x10²⁰

<h2>[Pb²⁺] = 2.31x10⁻²¹ M</h2>

Hope this helps

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7 0
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How many grams of zinc sulfide are used to produce 1.28 grams of zinc oxide?
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0.012288122055459

Explanation:

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2 years ago
Find an expression for the change in entropy when two blocks of the same substance of equal mass, one at the temperature Th and
Gre4nikov [31]

Explanation:

Relation between entropy change and specific heat is as follows.

            \Delta S = C_{p} log (\frac{T_{2}}{T_{1}})

The given data is as follows.

     mass = 500 g,         C_{p} = 24.4 J/mol K

     T_{h} = 500 K,          T_{c} = 250 K               

   Mass number of copper = 63.54 g /mol

Number of moles = \frac{mass}{/text{\molar mass}}

                                 = \frac{500}{63.54}

                                 = 7.86 moles

Now, equating the entropy change for both the substances as follows.

     7.86 \times 24.4 \times [T_{f} - 250] = 7.86 \times 24.4 \times [500 -T_{f}]

       T_{f} - 250 = 500 - T_{f}

          2T_{f} = 750

So,       T_{f} = 375^{o}C

  • For the metal block A,  change in entropy is as follows.

         \Delta S = C_{p} log (\frac{T_{2}}{T_{1}})

              = 24.4 log [\frac{375}{500}]

              = -3.04 J/ K mol

  • For the block B,  change in entropy is as follows.

         \Delta S = C_{p} log (\frac{T_{2}}{T_{1}})

                  = 24.4 log [\frac{375}{250}]

                  = 4.296  J/Kmol

And, total entropy change will be as follows.

                       = 4.296 + (-3.04)

                      = 1.256 J/Kmol

Thus, we can conclude that change in entropy of block A is -3.04 J/ K mol  and change in entropy of block B is 4.296  J/Kmol.

8 0
3 years ago
Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutral
Nuetrik [128]

Answer:

a. The maximum volume of 0.143 M HCl required is 154.4 mL.

b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

a.

Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O

Mass of aluminum hydroxide = 350 mg =  0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = \frac{0.350 g}{78 g/mol}=0.004487 mol

According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :

\frac{3}{1}\times 0.004487 mol=0.01346 mol of HCl.

Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O

Mass of magnesium hydroxide = 250 mg =  0.250 g ( 1mg = 0.001 g)

Moles of magnesium hydroxide = \frac{0.250 g}{58 g/mol}=0.004310 mol

According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310  mole of magnesium hydroxide will be neutralize by :

\frac{2}{1}\times 0.004310 mol=0.008621 mol of HCl.

Total moles of HCl required to neutralize both :

0.01346 mol + 0.008621 mol = 0.02208 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}

V=\frac{0.02208 mol}{0.143 M}=0.1544 L

1 L = 1000 mL

0.1544 L = 154.4 mL

The maximum volume of 0.143 M HCl required is 154.4 mL.

b.

CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2

Mass of calcium carbonate = 970mg =  0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = \frac{0.970 g}{100 g/mol}=0.00970 mol

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

\frac{2}{1}\times 0.00970 mol=0.0194 mol of HCl.

Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}

V=\frac{0.0194 mol}{0.143 M}=0.1357 L

1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

4 0
3 years ago
Someone pls help me I will make you brain
mariarad [96]

Answer:

B

Explanation:

7 0
3 years ago
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