问题描述：
英语翻译
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!
(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)
还有一部分,
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!
(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)
还有一部分,
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