How much NaOH is contained in 35.0 mL of a 1.10 M solution of NaOH in H2O?
回答 (3)
2018-01-26 10:12 am
Molarity of NaOH = (No. of moles of NaOH) × (Volume of the solution in L)
No. of moles of NaOH = (1.10 mol/L) × (35.0/1000 L) = 0.0385 mol
Molar mass of NaOH = (23.0 + 16.0 + 1.0) g/mol = 40.0 g/mol
Mass of NaOH = (0.0385 mol) × (40.0 g/mol) = 1.54 g
No. of moles of NaOH = (1.10 mol/L) × (35.0/1000 L) = 0.0385 mol
Molar mass of NaOH = (23.0 + 16.0 + 1.0) g/mol = 40.0 g/mol
Mass of NaOH = (0.0385 mol) × (40.0 g/mol) = 1.54 g
2018-01-27 6:45 pm
1.1 mol/L * 0.035 L = 0.0385 mol NaOH
MM(NaOH) = 40 g/mol
0.0385 mol x 40 g/mol = 1.54 g NaOH
MM(NaOH) = 40 g/mol
0.0385 mol x 40 g/mol = 1.54 g NaOH
2018-01-26 10:01 am
Moles: Volume times concentration -- so, 0.035 mL * 1.10 moles/L
Grams: moles times molecular weight -- so the answer from above multiplied by 40 g/mole
Grams: moles times molecular weight -- so the answer from above multiplied by 40 g/mole
收錄日期: 2021-04-18 18:06:33
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