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kakasveta [241]
3 years ago
11

How many mL of 14.5M lithium carbonate solution must be used to deliver 4.20 g of lithium ion

Chemistry
1 answer:
svet-max [94.6K]3 years ago
7 0

Answer:

10.3 g Li) / (6.9410 g Li/mol) x (1 mol Li3PO4 / 3 mol Li) / (0.750 mol/L Li3PO4) = 0.6595 L = 660. mL

Explanation:

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How many grams of fluorine are contained in 8 molecules of boron trifluoride?
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<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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<span>2PO43–(aq) + 3Ca2+(aq) → Ca3(PO4)2(s)
</span>
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