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Iteru [2.4K]
2 years ago
15

ZnS (s) + AIP(s) → AI₂S₂(s) + Zn P₂(s)

Chemistry
1 answer:
Allisa [31]2 years ago
3 0

Answer:

3 ZnS (s) + 2 AIP(s) ---> AI₂S₃(s) +  Zn₃P₂(s)

Explanation:

I believe you are asking for this reaction to be balanced. However, after looking more closely, those are not the products that would be created from the reactants. Assuming this is a double-displacement reaction, and taking the most common charges of the ions into consideration, the actual products would be AI₂S₃(s) + Zn₃P₂(s). I apologize if I assumed incorrectly.

For an equation to be balanced, there needs to be the same amount of each element on both sides of the equation.

The unbalanced equation:

ZnS (s) + AIP(s) ---> AI₂S₃(s) + Zn₃P₂(s)

<u>Reactants</u>: 1 zinc, 1 sulfur, 1 aluminum, 1 phosphorus

<u>Products</u>: 3 zinc, 3 sulfur, 2 aluminum, 2 phosphorus

As you can see, there is an unequal amount of each element on the reactants and products side. To balance the equation, coefficients in front of the compound(s) will be necessary.

The balanced equation:

3 ZnS (s) + 2 AIP(s) ---> AI₂S₃(s) +  Zn₃P₂(s)

<u>Reactants</u>: 3 zinc, 3 sulfur, 2 aluminum, 2 phosphorus

<u>Products</u>: 3 zinc, 3 sulfur, 2 aluminum, 2 phosphorus

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Most Bic lighters hold 5.0ml of liquified butane (density = 0.60 g/ml). Calculate the minimum size container you would need to "
Hatshy [7]

Answer:

Volume of container = 0.0012 m³ or 1.2 L or 1200 ml

Explanation:

Volume of butane = 5.0 ml

density = 0.60 g/ml

Room temperature (T) = 293.15 K

Normal pressure (P) = 1 atm = 101,325 pa

Ideal gas constant (R) = 8.3145 J/mole.K)

volume of container V = ?

Solution

To find out the volume of container we use ideal gas equation

PV = nRT

P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

First we find out number of moles

<em>As Mass = density × volume</em>

mass of butane = 0.60 g/ml ×5.0 ml

mass of butane = 3 g

now find out number of moles (n)

n = mass / molar mass

n = 3 g / 58.12 g/mol

n = 0.05 mol

Now put all values in ideal gas equation

<em>PV = nRt</em>

<em>V = nRT/P</em>

V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa

V = 121.87 ÷ 101,325 pa

V = 0.0012 m³ OR 1.2 L OR 1200 ml

8 0
4 years ago
What is the overall cell potential for this redox reaction?
alexgriva [62]
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Why could weathering and erosion be a bad thing?
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The can lead to mass wasting

Explanation:

Weathering and erosion can be a bad thing because they can lead or trigger mass wasting.

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  • Both weathering and erosion can lead to serious problems like mass-wasting because they are earth moving processes.
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3 0
4 years ago
1. write a sentence that describes how the food changes in the stomach.
Advocard [28]

Explanation:

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8 0
3 years ago
Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant
antoniya [11.8K]

Answer: The value of E^{o} for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

          AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)

Its solubility product will be as follows.

       K_{sp} = [Ag^{+}][Cl^{-}]

Cell reaction for this equation is as follows.

     Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)

Reduction half-reaction: Ag^{+} + 1e^{-} \rightarrow Ag(s),  E^{o}_{Ag^{+}/Ag} = 0.799 V

Oxidation half-reaction: Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-},   E^{o}_{AgCl/Ag} = ?

Cell reaction: Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)

So, for this cell reaction the number of moles of electrons transferred are n = 1.

    Solubility product, K_{sp} = [Ag^{+}][Cl^{-}]

                                               = 1.77 \times 10^{-10}

Therefore, according to the Nernst equation

           E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

At equilibrium, E_{cell} = 0.00 V

Putting the given values into the above formula as follows.

         E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

        0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}    

       E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}

                  = 0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}

                  = 0.577 V

Hence, we will calculate the standard cell potential as follows.

           E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}

       0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}

       0.577 V = 0.799 V - E^{o}_{AgCl/Ag}

       E^{o}_{AgCl/Ag} = 0.222 V

Thus, we can conclude that value of E^{o} for the half-cell reaction is 0.222 V.

3 0
3 years ago
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