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1. ### concentration calculation

Suppose that your question refers to % by mass, then: In 1 dm3 of 2 M NaCl solution: No. of moles of NaCl = 2 with mass = 2 x 58.5 = 117 g Mass of water = 1000 g So % by mass of NaCl = 117/(117 + 1000) x 100% = 10.47%

分類：科學及數學 > 化學 2008年03月20日

2. ### measurment calculation

Volume * density = mass Volume = mass / density = 23 / 7.86 cm^3 = 2.9262 cm^3 = 2.9262 mm^3 * 10^3 = 2926.2 mm^3

分類：科學及數學 > 化學 2007年09月27日

3. ### F4 calculation

1. NH3 + 2O2 → HNO3 + H2O 2. HNO3 + NH3 → NH4NO3 According to the equations, we should use 8.5 tonnes of ammonia in state one and the other 8.5 tonnes in state 2. (By mole ratio, NH3 : HNO3...

分類：科學及數學 > 化學 2007年04月19日

4. ### chemical calculation ...

Let 1:y be the mole ratio of X to oxygen in the oxide [ (16*y) / (16*y + 74.9) ] x 100% = 24.3% 16y = 3.888 + 18.2007 y = 1.38 ( corr. to 3sig. fig.) 1 : 1.38 ~ 2 : 3 the mole ratio of X to oxygen in the oxide is 2:3

5. ### chemical calculation

首先,解條題目: 34g的zine suplphate 於75cm^3的水內,那此溶液的濃度是多少? in &#39;M&#39; = 濃度單位 moles per cubic decimeter = mol dm ^-3 = Molar 統稱 到高程度時,濃度= M, uM, nM etc...M=1000uM 等. 但你這裡唔駛用. 首先知道zine suplphate...

分類：科學及數學 > 化學 2007年03月24日

6. ### Titraion... calculation

no. of mole of EDTA = (27.32/1000)x0.00396 = 1.08187X10^-5 no. of mole of Mg2+ = 1.08187X10^-5 no. of mole of Mg2+ in the 2L sample = 1.08187X10^-5 X (2000/10) = 2.13X10^-3 Conc. = 2.13X10^-3 / 2 = 1.082X10^-3...

分類：科學及數學 > 化學 2008年07月16日

7. ### Calculation Problem

分類：科學及數學 > 數學 2014年05月16日

8. ### calculation

You assumed thatthe monoprotic acid is completely dissociated to give the hydrogen ions. However, the pH value shows that it is not completely dissociated. pH = -(Log[H ]), so 2.69 = -(log [H ]) and...

分類：科學及數學 > 化學 2008年05月25日

9. ### calculation

Since when it is cooled down to the room temperature, the product has been condensed to liquid water. According to the given, no. of moles of water produced = 0.00083 moles which is equivalent to 0.015 g of water, i.e. 0...

分類：科學及數學 > 化學 2007年05月31日

10. ### calculation -heat

(c) The heat gain by the dripping water in 1 sec = (20/1000)x 4200 x T where T is the rise in temperature of the water, and 4200 J/kg-K is the specific heat capacity of water Heat generated by the drill = 2500 x 0.6 J/s The temperature of the...