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July 28, 2014

July 28, 2014

Posted by **tatianna** on Tuesday, September 3, 2013 at 1:29am.

- chemistry -
**Graham**, Tuesday, September 3, 2013 at 3:49amThe ratio of Mg to F is constant.

f2 / m2 = f1 / m1

To find the second mass of fluorine (f2),

rearrange and substitute appropriate values.

- chemistry -
**Steve**, Tuesday, September 3, 2013 at 3:49amF/Mg = 2.58/1.65

So, given 1.28kg Mg, we should have

1.28*(2.58/1.65) = 2.00kg F

- chemistry -
**Devron**, Tuesday, September 3, 2013 at 4:01amYou need to create an identity to solve for the mass of fluorine:

1.65 kg of Mg=2.58 kg of F

and

1.38 kg of Mg=x kg of F

So,

x kg of F/1.38kg of Mg=2.58 kg of F/1.65 kg of Mg

Solve for x kg of F,

x kg of F=(2.58 kg of F/1.65 kg of Mg)*1.38kg of Mg

x kg of F=2.157Kg of F

2.157 kg of F*(10^3g/1kg )=2.16 x 10^3 g of F or 2,160 g of F

Answer contains 3 sig figs.

- chemistry -
**Devron**, Tuesday, September 3, 2013 at 4:01amYou need to create an identity to solve for the mass of fluorine:

1.65 kg of Mg=2.58 kg of F

and

1.38 kg of Mg=x kg of F

So,

x kg of F/1.38kg of Mg=2.58 kg of F/1.65 kg of Mg

Solve for x kg of F,

x kg of F=(2.58 kg of F/1.65 kg of Mg)*1.38kg of Mg

x kg of F=2.157Kg of F

2.157 kg of F*(10^3g/1kg )=2.16 x 10^3 g of F or 2,160 g of F

Answer contains 3 sig figs.

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