Question

求高手把英文翻译成中文,不要机译,谢谢!

只要翻译文字部分就好,化学式和公式不用翻译,好的话继续加分,辛苦了!先谢谢了!!
(1) Calculate molecular proportions of the oxides by dividing by their molecular weights.
(2) Add MnO to FeO.
(3) Allocate CaO equal to 3.33 x P2O5 to apatite.
(4) If FeO > TiO2: allocate FeO equal to the amount of TiO2 present to ilmenite.If FeO < TiO2: an excess of Tio2 is provisionally made into sphene, using an equal amount of CaO (although, only after CaO has been allocated to anorthite). If there is still an excess of TiO2 it is allocated to rutile.
(5) Provisionally allocate Al2O3 equal to K2O for orthoclase.
(6) Provisionally allocate to any excess Al2O3 equal Na2O for albite. If there is insufficient Al203 go to step 10.
(7) Any excess of Al2O3 over Na2O + K2O Is matched with an equal amount of CaO for anorthite .
(8) If there Is an excess of Al2O3 over CaO it is allocated to corundum.
(9) An excess of CaO over Al203 is used for diopside and wollastonite.
(10) An excess of Na2O over Al203 is used in acmite, there Is no anorthite in the norm. Allocate Fe2O3 equal to the excess Na2O for acmite.
(11) If Fe2O3 > Na2O, allocate an equal amount of FeO for magnetite.
(12) If Fe2O3 is still in excess, it is calculated as hematite.
(13) Sum MgO + remaining FaO. Calculate their relative proportions.
(14) Any CaO unused after anorthite (step 7) is allocated to diopside using an equal amount of FeO + MgO (allotted in proportion to that determined in step 13).
(15) Excess CaO is provisionally allocated to wollastonite.
(16) Excess MgO + FeO Is provisionally allocated to hypersthene.
(17) Allocate SiO2 to sphene, acmite, provisional orthoclase,albite and anorthite,diopside, wollastonite and hypersthene in the proportions of the formulae above.
(18) An excess of SiO2 Is calculated as quartz.
(19)If there is insufficient SiO2 at step 17 the SiO2 allocated to hypersthene is omitted from the sum of SiO2 used. If at this stage there is an excess of SiO2, that remaining is allocated between hypersthene and olivine using the equations:
X=2S-M Y=M-x．
where x is the number of hypersthene molecules and y the number of olivine molecules, M the value of available MgO＋FeO and S the amount of available SiO2. If there is insufficient SiO2 to match half the amount of MgO+FeO, then MgO+FeO is made into olivine (rather than hypersthenel)
http://zhidao.baidu.com/question/124226388.html
还有一部分,谢谢了

Answer 1

1）计算由其分子量分化的氧化物分子的比例。
（2）添加二氧化锰的矿FeO。
（3）分配钙等于3.33 x五氧化二磷的磷灰石。
（4如果矿FeO“二氧化钛）：分配中FeO等于二氧化钛在场ilmenite.If矿FeO量”二氧化钛：一个是暂时过剩二氧化钛制成榍石，使用等量的钙（虽然，只有在已分配钙以钙长石）。如果仍然过剩的二氧化钛是分配给金红石。
（5）氧化铝平等分配暂定为钾长石钾。
（6）具临时分配给任何多余的钠长石氧化铝平等氧化钠。如果没有足够的Al203转到步骤10。
（7）任何超过氧化钠过量氧化铝+ K2O为与氧化钙的钙长石等额匹配。
（8）如果有一个以上的Al2O3氧化钙是分配给刚玉过剩。
（9）对CaO对Al203过剩用于透辉石，硅灰石。
（10）一个氧化钠对Al203过剩用于鉥，没有任何规范钙长石。分配氧化铁等于为鉥过量氧化钠。
（11）如氧化铁“氧化钠，拨出的磁铁矿中FeO同等的数额。
（12）如氧化铁仍然过剩，计算公式为赤铁矿。
（13）心氧化镁+其余粮农组织。计算它们的相对比例。
（14）后钙长石（氧化钙未使用的步骤7）透辉石，是分配给使用同等数量的FeO +氧化镁（分配比例，在步骤13确定）。
（15）过量钙是暂时分配给硅灰石。
（16）过量MgO +中FeO被临时分配给苏辉。
（17）分配二氧化硅的榍石，鉥，临时钾长石，钠长石和钙长石，透辉石，硅灰石和苏辉在上述公式的比例。
（18）二氧化硅过剩的一个计算公式为石英。
（19）如果在第17步分配给苏辉二氧化硅不足的二氧化硅省略由二氧化硅的款项。如果在这个阶段有一个多余的SiO2，即余下的则是橄榄石之间苏辉和分配使用的公式：
X = 2秒，张敏=的M -的X。
其中x是苏辉分子数和y的橄榄石分子，男可用氧化镁值数+ FeO和S的可用二氧化硅量。如果没有足够的二氧化硅匹配量的一半氧化镁+矿FeO，然后氧化镁+分为橄榄石中FeO（而不是hypersthenel制）
QQ爱好者为你服务

Answer 2

下载谷歌金山词霸合作版，一定可以翻译你的句子的。我也是用这个的！建议你用吧！！！！！！

Answer 3

分太少了....太麻烦了

Answer 4

20分，看都懒得看完了。