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15.6吋 可側背 手提 防震包 (DH120)

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15.6吋 可側背 手提 防震包 (DH120)

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15.6吋 可側背 手提 防震包 (DH120)

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標題:

高中化學問題 幫忙一下

發問:

1.What is molarity of Na+ in a solution of NaCl(AW of Na:23g，AW of Cl:35g) whose salinity(grams of NaCl in 1kg of seawater) is 5.8，if thesolution has a density of 1g/ml？Molarity(M) is defined as the number of moles of solute in a liter of solution(mole/liter)2.estimate the PH of 2.0M solution of... 顯示更多 1.What is molarity of Na+ in a solution of NaCl(AW of Na:23g，AW of Cl:35g) whose salinity(grams of NaCl in 1kg of seawater) is 5.8，if thesolution has a density of 1g/ml？Molarity(M) is defined as the number of moles of solute in a liter of solution(mole/liter) 2.estimate the PH of 2.0M solution of H2SO4 列個算式解釋一下 謝謝

最佳解答:

1.設取1kg(=1000g)海水， 體積V=1000g÷1.0(a density of 1g/ml)=1000ml=1 L NaCl有5.8g => 5.8/(23+35.5) = 0.099mol NaCl(aq)→Na+(aq) +Cl-(aq) => 解離生成Na+ = 0.099 mol Thus,the molarity of Na+ [Na+]= 0.099mol/1 L =0.099M≒0.1M 2.[H2SO4]=2M H2SO4→2H+ + SO4(2-) => 解離生成[H+] = 4M pH= - log[H+] = -log4 = -2log2= - 0.602

其他解答:

H2SO4 is a strong acid, but bisulfate ion (HSO4(-)) ion is not. One has to consider the reversible nature of ionization of such ion ---- pH should be higher than that you've calculated.

15.6吋 可側背 手提 防震包 (DH120)

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