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AlexFokin [52]
3 years ago
14

How do you balance Sb+I2=SbI3 show work please

Chemistry
1 answer:
egoroff_w [7]3 years ago
7 0

Answer:

 2Sb   +    3I₂    →    2SbI₃

Explanation:

The reaction expression we need to balance is:

         Sb   +    I₂    →    SbI₃

To balance this expression, we need to assign the variables a,b and c to each specie as the coefficient that will balance the expression.

              aSb   +    bI₂    →    cSbI₃

Conserving Sb :  a  = c

                      I :   2b  = 3c

let a  = 1, c  = 1,  b  = \frac{3}{2}  

   Multiply through by 2;

 a  = 2, b  = 3  and c  = 2

         2Sb   +    3I₂    →    2SbI₃

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3 years ago
Read 2 more answers
A 23.5g aluminum block is warmed to 65.9°C and plunged into an insulated beaker containing 55.0g water initially at 22.3°C. The
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Answer:

25.97oC

Explanation:

Heat lost by aluminum = heat gained by water

M(Al) x C(Al) x [ Temp(Al) – Temp(Al+H2O) ] = M(H2O) x C(H2O) x [ Temp(Al+H2O) – Temp(H2O) ]

Where M(Al) = 23.5g, C(Al) = specific heat capacity of aluminum = 0.900J/goC, Temp(Al) = 65.9oC, Temp(Al+H2O)= temperature of water and aluminum at equilibrium = ?, M(H2O) = 55.0g, C(H2O)= specific heat capacity of liquid water = 4.186J/goC

Let Temp(Al+H2O) = X

23.5 x 0.900 x (65.9-X) = 55.0 x 4.186 x (X-22.3)

21.15(65.9-X) = 230.23(X-22.3)

1393.785 - 21.15X = 230.23X – 5134.129

230.23X + 21.15X = 1393.785 + 5134.129

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X = 6527.909/251.38

X = 25.97oC

So, the final temperature of the water and aluminum is = 25.97oC

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3 years ago
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