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Xelga [282]
4 years ago
12

What is the answer to this chemistry question?

Chemistry
1 answer:
olga_2 [115]4 years ago
5 0

Answer:

The answer to your question is remplacement double

Explanation:

Data

Lead (II) nitrate = Pb(NO₃)₂

Potassium iodide = KI

Process

1.- Write the balanced chemical reaction

             Pb(NO₃)₂  +  2KI  ⇒   PbI₂  +  2KNO₃

2.- Conclusion

This is a remplacement double reaction because there are two reactants that interchange cations and the products are a combination of the reactants.

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Problem PageQuestion Iron(III) oxide and hydrogen react to form iron and water, like this: (s)(g)(s)(g) At a certain temperature
Marrrta [24]

Complete question

The complete question is shown on the first uploaded image

Answer:

The value is  K_c  =  2.69 *10^{-5}  

Explanation:

From the question we are told that

   The equation is  

           Fe_20_3_{(s)}+3H_{(g)}\to2Fe_{(s)}+3H_2O_{(g)}

Generally the equilibrium is mathematically represented as

        K_c  =  \frac{[H_2O]^2}{[H_2]^3}

Here [H_2O] is the concentration of water vapor which is mathematically represented as

      [H_2O ] =  \frac{n_w}{V_s }

Here V_s is the volume of the solution given as 8.9 L

n_w is the number of moles of water vapor which is mathematically represented as

        n_w  =  \frac{m_w}{Z_w}

Here  m_w  is the mass of water given as 2.00 g

and   Z_w  is the molar mass of water with value  18 g/mol

So  

         n_w  =  \frac{2}{18}

=>     n_w  = 0.11 \  mol

So

     [H_2O ] =  \frac{0.11}{8.9 }

=>   [H_2O ] = 0.01236 \  M

Also

[H] is the concentration of hydrogen gas which is mathematically represented as

      [H ] =  \frac{n_v}{V_s }

Here V_s is the volume of the solution given as 8.9 L

n_v is the number of moles of  hydrogen gas which is mathematically represented as

        n_v  =  \frac{m_v}{Z_v}

Here  m_w  is the mass of water given as 4.77 g

and   Z_v  is the molar mass of water with value  2 g/mol

So  

         n_w  =  \frac{4.77}{2}

=>     n_w  = 2.385 \  mol

So

     [H_2O ] =  \frac{2.385}{8.9 }

=>   [H_2O ] =  0.265 \  M

So

     K_c  =  \frac{( 0.01236 )^3}{ (0.265 )^2}

=>   K_c  =  2.69 *10^{-5}  

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