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Ber [7]
3 years ago
15

The compound magnesium iodide is a strong electrolyte write the reaction when solid magnesium iodide is put into water

Chemistry
1 answer:
mestny [16]3 years ago
7 0
  The  reaction  for magnesium  iodide  when put into water  is  as below

 MgI2(s) →  Mg^2+(aq)   + 2I^-(aq)

when  magnesium iodide  but into water it dissociate/ ionize  completely  into  Mg^2+  and  2l^-   ions. Magnesium iodide dissociate/ionize  completely because  magnesium  iodide  is a strong electrolyte  which  dissociate/ ionize  completely into their ions  when it is put into   water .
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Answer:

Volume of CHCl_3   = 15.31 mL

Volume of CHBr_3 = 4.69 mL

Explanation:

Given that:

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Suppose the volume of  CHCl_3 = P ml

and the volume of CHBr_3 = Q ml

the sum of their volumes should be equal to the total volume of the mixture

P \ ml + Q \ ml = 20 ml  ----- (1)

However, we know that Density = mass/volume

∴ mass = density × volume

The equation can now be expressed as:

\mathtt{(Density \ of  \ CHCl_3 \times Vol. \ of  \ CHCl_3 ) + (Density  \  of \  CHBr_3  \times \ volume \ of \ CHBr_3)} = \mathtt{ (Density  \ of \ mixture \times volume \ of \ the \ mixture)}

1.492 g/mL × P mL + 2.890 g/mL × Q mL = 1.82 g/mL × 20 mL  ---- (2)

From equation (1) ;

let Q = 20 - P

The replace the value of P into equation (2)

1.492 g/mL × P mL + 2.890g/mL × (20 - P) mL = 1.82 g/mL × 20 mL

1.492 P g + 57.8g - 2.890 P g =  36.4g

1.492 P g - 2.890 P g = 36.4g - 57.8g

-1.398 P g = -21.4g

P = -21.4g/-1.398g

P = 15.31 mL

Q = 20 - P

Q = (20 - 15.31) mL

Q = 4.69 mL

∴

Volume of CHCl_3   = 15.31 mL

Volume of CHBr_3 = 4.69 mL

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