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lisov135 [29]
4 years ago
5

You need to make an aqueous solution of 0.216 m manganese(ii) sulfate for an experiment in lab, using a 125 ml volumetric flask.

how much solid manganese(ii) sulfate should you add?
Chemistry
1 answer:
olya-2409 [2.1K]4 years ago
8 0
125 mL= 0.125 L
0.216 M==0.126 mol/L

0.126 mol/L*0.125L=0.0158 mol MnSO4 are needed
Molar mass (MnSO4)= 54.9+32.1+4*16.0= 151 g/mol

 151 g/mol*0.0158 mol=2.39 g solid  MnSO4
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Molar mass of C₆H₆O₃ = 126 g/mol

mass of C₆H₆O₃

                 mass in grams = no. of moles x molar mass

                 mass of C₆H₆O₃ = 1 mol x 126 g/mol

                 mass of C₆H₆O₃ = 126 g

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mass of O₂

                mass in grams = no. of moles x molar mass

                mass of O₂ = 3 mol x 32 g/mol

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96 g of O₂ combine with 126 g of C₆H₆O₃ then how many grams of C₆H₆O₃ will combine with 97.0grams of O2

Apply unity Formula

                      96 g of O₂ ≅ 126 g of C₆H₆O₃

                      97 g of O₂ ≅  X g of C₆H₆O₃

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                    X g of O₂ = 126 g x 97 g / 96 g

                   X g  of O₂ = 127.3 g

127.3 g of C₆H₆O₃ will combine with 97 g of O₂

So

mass of C₆H₆O =  127.3 g

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