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Leviafan [203]
3 years ago
6

Ferrous sulfate (FeSO4) was used in the

Chemistry
1 answer:
BARSIC [14]3 years ago
6 0

Ferrous sulfate (FeSO4) is obtained by the reaction of aqueous ferric sulfate with gaseous sulfur dioxide and water according to the equation below:

Aqueous ferric sulfate = Fe_2(SO_4)_3_{(aq)}

Gaseous sulfur dioxide = SO_2_{(g)}

Water = H_2O_{(l)}

Aqueous sulfuric acid =  H_2SO_4_{ (aq)}

Aqueous ferrous sulfate =  FeSO_4_{ (aq)}

Thus, the balance equation becomes:

Fe_2(SO_4)_3_{(aq)} +  SO_2_{(g)} + 2H_2O_{(l)} ---> 2 H_2SO_4_{ (aq)} + 2 FeSO_4_{ (aq)}

More on ferrous sulfate can be found here:  brainly.com/question/10834028

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A chemistry student needs of acetone for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student dis
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Answer:

The answer is "7.90 \ g".

Explanation:

Please finds the complete question in the attached file.

Volume = 10.0 \ mL \\\\Density =0.79 g\  cm^3

\text{Density}=\frac{Mass}{Volume}

Mass = Density \times Volume

         =(0.790 g\ cm^3) \times 10.0 \ mL\\\\=7.90 \ g

The mass of acetone to be weighed is 7.90 \ g

8 0
3 years ago
How many grams of fluorine are contained in 8 molecules of boron trifluoride?
Lelu [443]
<h3>Answer:</h3>

             7.57 × 10⁻²² g of F

<h3>Solution:</h3>

Data Given:

                 Number of Molecules  =  8

                 M.Mass of BF₃ =  67.82 g.mol⁻¹

                 Mass of Fluorine atoms  =  ?

Step 1: Calculate Moles of BF₃

           Moles  =  Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Putting value,

            Moles  =   8 Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

            Moles  =  1.33 × 10⁻²³ mol

Step 2: Calculate Mass of BF₃:

                   Moles  =  Mass ÷ M.Mass

Solving for Mass,

                   Mass  =  Moles × M.Mass

Putting values,

                   Mass  =  1.33 × 10⁻²³ mol × 67.82 g.mol⁻¹

                   Mass  =  9.0 × 10⁻²² g

Step 3: Calculate Mass of Fluorine Atoms:

As,

                         67.82 g BF₃ contains  =  57 g of F

So,

                    9.0 × 10⁻²² g will contain  =  X g of F

Solving for X,

                       X =  (9.0 × 10⁻²² g × 57 g) ÷ 67.82 g

                        X  =  7.57 × 10⁻²² g of F

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4 years ago
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