34.62 mL of 0.1510 M NaOH was needed to neutralize 50.0 mL of an H2SO4 solution. What is the concentration of the original sulfuric acid solution?
A) 0.0229 M B) 0.218 M C) 0.0523 M D) 0.209 M E) 0.105 M
Lithium metal dissolves in water to yield hydrogen gas and aqueous lithium hydroxide. What is the final concentration of hydroxide ions when 5.500 g of lithium metal is dropped into 750. mL of water?
A) 1.06 M B) 0.528 M C) 2.11 M D) 0.792 M E) 0.943 M
A 250. mL sample of 0.0328M HCl is partially neutralized by the addition of 100. mL of 0.0245M NaOH. Find the concentration of hydrochloric acid in the resulting solution.
A) 0.00700 M B) 0.0164 M C) 0.0383 M D) 0.0230 M E) 0.0575 M
1.40 g of silver nitrate is dissolved in 125 mL of water. To this solution 5.00 mL of 1.50M hydrochloric acid is added, and a precipitate forms. Find the concentration of silver ions remaining in solution.
A)5.7×10–3 MD)0.121M
B)6.34×10–2 ME)5.9×10–3 M
C)5.77×10–2 M
不好意思我還想問大大這幾題^^謝謝
化學 關於體積莫爾濃度
2012-11-11 9:58 pm
回答 (1)
2012-11-12 8:37 pm
✔ 最佳答案
1.The answer is : C) 0.0523 M
2NaOH + H2SO4 → Na2SO4+ 2H2O
No. of moles of NaOH = 0.1510 x (34.62/1000) = 0.005228 mol
No. of moles of H2SO4 = 0.005228 x (1/2) = 0.002614 mol
Concentration of H2SO4 = 0.002614 / (50/1000) = 0.0523 M
2.
The answer is : A) 1.06 M
2Li + 2H2O → 2Li⁺ + 2OH⁻ + H2
No. of moles of Li = 5.500/6.9 = 0.797 mol
No. of moles of OH⁻ ions = 0.797 mol
Concentration of OH⁻ ions = 0.797 / (750/1000) = 1.06 M
3.
The answer is : B) 0.0164 M
HCl + NaOH → NaCl + H2O
Initial no. of moles of HCl = 0.0328 x (250/1000) = 0.0082 mol
Initial no. of moles of NaOH = 0.0245 x (100/1000) = 0.00245 mol
NaOH is the limiting reactant (completely reacted), and HCl is in excess.
No. of moles of NaOH reacted = 0.0245 x (100/1000) = 0.00245 mol
No. of moles of HCl reacted = 0.00245 mol
No. of moles of HCl unreacted = 0.0082 - 0.00245 = 0.00575 mol
Volume of final solution = (250 + 100)/1000 = 0.35 L
Concentration of HCl in the resulting solution = 0.00575/0.35 = 0.0164 M
4.
The answer is : A) 5.7 x 10⁻³ M
AgNO3(aq) + HCl(aq) → AgCl(s) + H2O(l)
Initial number of moles of AgNO3 = 1.40/(107.9 + 14 + 16x3) =0.00824 mol
Initial number of moles of HCl = 1.50 x (5/1000) = 0.0075 mol
HCl is the limiting reactant (completely reacted), and AgNO3 is inexcess.
No. of moles of HCl reacted = 0.0075 mol
No. of moles of AgNO3 reacted = 0.0075 mol
No. of moles of AgNO3 unreacted = 0.00824 - 0.0075 = 0.00074 mol
Volume of final solution = (125 + 5)/1000 = 0.13 L
Concentration of AgNO3 remaining in solution = 0.00074/0.13 = 5.7 x10⁻³ M
Concentration of Ag⁺ ions remaining in solution = 5.7 x 10⁻³ M
參考: 賣女孩的火柴
收錄日期: 2021-04-13 19:05:53
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20121111000015KK03466