After first dilution,
After second dilution,
M3=(3.95×10−5)(0.002)0.010=7.90×10−6 MM_3=\frac{(3.95\times10^{-5})(0.002)}{0.010}=7.90\times10^{-6}\ MM3=0.010(3.95×10−5)(0.002)=7.90×10−6 M
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Dear maryam albalushi post a new task
1.00 mL of a 3.50 × 10–4 M solution of oleic acid is diluted with 9.00 mL of petroleum ether, forming solution A. 2.00 mL of solution A is diluted with 8.00 mL of petroleum ether, forming solution B. How many grams of oleic acid is 5.00 mL of solution B? (molar mass for oleic acid = 282 g/mol)
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Dear maryam albalushi post a new task
1.00 mL of a 3.50 × 10–4 M solution of oleic acid is diluted with 9.00 mL of petroleum ether, forming solution A. 2.00 mL of solution A is diluted with 8.00 mL of petroleum ether, forming solution B. How many grams of oleic acid is 5.00 mL of solution B? (molar mass for oleic acid = 282 g/mol)
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