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  1. Consider the original alkali: pH = 12 pOH = 14 - pH = 14 - 12 = 2 [OH-] = 10-pOH = 10-2 = 0.01 M Consider the diluted solution: It is diluted into 2%. [OH-] = 0.01 x 2% = 0.0002 M pOH = -log...

    分類:科學及數學 > 化學 2008年06月19日

  2. CPI in 1983= CPI in the base year CPI in 1960=CPI in the current year The price of a gallon of petrol in 1960 in 1983 =current price * CPI in the base year/CPI in current year =US$0.31*100/30 =US$1.0333333... =US$1.03

  3. 因爲你淨係要方程式,無需要數字,如果你有比數字,我可以幫你計,如果你要我幫你計,vr29w2hl@yahoo.com.hk 方程式如下 : 第一步 先將R1同R2以並聯方式加起來=Ra 再將R3同R4以並聯方式加起來=Rb 接下來將Ra同Rb以串聯方式...

  4. As follows: 圖片參考:http://i117.photobucket.com/albums/o61/billy_hywung/Sep07/Crazydiff28.jpg

    分類:科學及數學 > 化學 2007年09月18日

  5. For gases at the same temperature and pressure, volume ratio = mole ratio. 2H2(g) + O2(g) → 2H2O(g) Volume ratio H2 : O2 = 2 : 1 Volume of H2 = 100 x 49% = 49 cm3 Volume of O2 needed = 49/2 = 24.5 cm3...

    分類:科學及數學 > 化學 2007年09月11日

  6. mass no. Al=27.0 S=32.1 O=16.0 H=1.0 Avogadro constant = L = 6.02*10^23 formula mass of hydrated aluminium sulphate=27*2+(32.1+16*4)*3+6*(1*2+16)=450.3 no. of moles in sample = 67.5/450.3 = 0.15 mole - the no. of moles of...

    分類:科學及數學 > 化學 2007年01月31日

  7. d(f(4x^2))/dx = 9x^2 [d(f(4x^2))/d(4x^2)][d(4x^2)/dx] = 9x^2 [d(f(4x^2))/d(4x^2)][8x] = 9x^2 [d(f(4x^2))/d(4x^2)] = (9/8)x Let y = 4x^2 d(f(y))/dy = (9/8)[(√y)/2] = (9√y)/16 But x or y are just symbols. therefore f'(x) = (9√x)/16

    分類:科學及數學 > 數學 2013年10月22日

  8. Taking 1 litre of such solution, then there are 984 g in which 100 g are ethanol and 884 g are water. So for (1), we have: 116 cm3 of ethanol weighs 100 g (Since vol. of water = 884 cm3) No. of moles of ethanol = 100/46 = 2.174 Hence molarity...

    分類:科學及數學 > 化學 2009年03月27日

  9. Let r (cm) be the radius of the sphere. Volume of water=(pi)(23^2)(27)=14283 pi (cm^3) Because the sphere is just immersed in the water,the height of water is 2r (cm). (pi)(4/3)(r^3)+14283(pi) = (pi)(23^2)(2r) 4(r^3)-3174r + 42849 = 0 r = -33.38253764 ∴There...

    分類:科學及數學 > 數學 2010年04月03日

  10. (commercial vinegar) 1. Consider the titration: CH3COOH + NaOH → CH3COONa + H2O No. of moles of NaOH used = MV = 0.2 x (20/1000) = 0.004 mol No. of moles of CH3COOH = 0.004 mol Volume of...

    分類:科學及數學 > 化學 2008年05月04日