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san4es73 [151]
2 years ago
4

What mass of magnesium bromide would be required to prepare 720?

Chemistry
1 answer:
fomenos2 years ago
5 0
Complete Question:
                               <span>What mass of magnesium bromide would be required to prepare 720mL of a 0.0939M aqueous solution?

Answer:
            Molarity is calculated using following formula,

                               Molarity  =  Moles / Volume of Solution  ---- (1)

Data Give;
                 Molarity  =  </span>0.0939 mol.L⁻¹

                 Volume  =  720 mL  =  0.72 L

Solving eq.1 for Moles,

                            Moles  =  Molarity × Volume

                            Moles  =  0.0939 mol.L⁻¹ × 0.72 L

                            Moles  =  0.0676 moles

Now convert Moles into Mass using following formula,

                            Moles  =  Mass / M.Mass

Solving for Mass,

                            Mass  =  Moles × M.Mass

                            Mass  =  0.0676 mol × 184.113 g.mol⁻¹

                            Mass  =  12.44 grams
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Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N
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The balanced equilibrium reaction is,

                             N_2O_4(g)\rightleftharpoons 2NO_2(aq)

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At eqm. conc.     (1.0-x) M    (2x) M

As we are given,

The percent of dissociation of N_2O_4 = \alpha = 28.0 %

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