How many moles of Ba(OH)2 are present in 105 mL of 0.200 M Ba(OH)2?
回答 (2)
2018-06-08 12:15 am
✔ 最佳答案
No. of moles of Ba(OH)₂ = (0.200 mol/L) × (105/1000 L) = 0.0210 mol
2018-06-08 3:11 am
Remember the eq'n
moles = [concc'n] x vol(mL) / 1000
mol(Ba(OH)2 = 0.2 x 105 / 1000
mol(Ba(OH)2 = 0.021
moles = [concc'n] x vol(mL) / 1000
mol(Ba(OH)2 = 0.2 x 105 / 1000
mol(Ba(OH)2 = 0.021
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