Question

求助帮满翻译一下化学英文文章，要通顺的，谢谢了。

大一老师要我们化学翻译和托福一样的文章，翻译的不通顺还不行，太坑爹了。
一点分不多，真心求助。谢谢。
4.4.2 Residence Times within Individual Ocean basins
Due to the very uneven geographic distribution of the wet deposition of dissolved Fe,and the very uneven distribution of export production and its ensuing mobilisation from surface sediments,there is not much validity in a global residence time concept for an element such as Fe.Moreover,the net river input,albeit modest compared with the other sources,is also very unevenly distributed with major inputs into the central Atlantic Ocean.
thus,a more valid approach would be to construct mass balances for each ocean basin.for example ,the north and equatorial Atlantic Ocean from~10 S to 50 N receives one third or 3-8x109moly-1 dissoved Fe from aeolian input.adding to this one half of the global river input of Fe,or 1.3x109 moly-1,the combined inputs add up to 4.3-9.3x109 moll-1.the upper 100m depth has a volume of 3x1015m3,and with a mean Fe concentration of 0.2nmoll-1,a dissolved Fe inventory of 0.6x109mol.the surface water residence time then calculates as 0.07-0.15years or only 1-2 months.this would be even shorter when Fe from sedimentary sources is included.simply exporting all this Fe towards the seafloor,and again assuming some 50% mobilisation from deep sea sediments,the estimated input of dissolved Fe into deep waters becomes 2.0-4.5x109 moly-1.Given the total volume of this basin,and an assumed deep water concentration of 1nmoll-1,the Fe inventory is 100x109mol and the estimated residence time is 22-50years.

Answer 1

4.4.2 Residence Times within Individual Ocean basins
单个海洋盆地中的的“停留时间”
Due to the very uneven geographic distribution of the wet deposition of
由于溶解的铁的湿沉淀物在地理上分布十分不均衡和输出物的不均分配导致的表面沉积物的活动，对于铁这样的元素，在全球范围内并没有一个固定的“停留时间（这玩意你自己找术语去）”概念。
dissolved Fe,and the very uneven distribution of export production and its ensuing mobilisation from surface sediments,there is not much validity in a global residence time concept for an element such as Fe.
并且，净河输入（啥玩意），尽管这玩意与其他资源相比已经十分均衡，但是由于主要向大西洋中心输入，本身依然存在分配不均的问题。Moreover,the net river input,albeit modest compared with the other sources,is also very unevenly distributed with major inputs into the central Atlantic Ocean.
这就是说，一个更加有效的方式将会构架起各个海洋盆地的总体平衡。
thus,a more valid approach would be to construct mass balances for each ocean basin.
我觉得下面这部分你还是自己看吧，英语部分要是看不懂你可以继续复习高中英语了= =
for example ,the north and equatorial Atlantic Ocean from~10 S to 50 N receives one third or 3-8x109moly-1 dissoved Fe from aeolian input.adding to this one half of the global river input of Fe,or 1.3x109 moly-1,the combined inputs add up to 4.3-9.3x109 moll-1.the upper 100m depth has a volume of 3x1015m3,and with a mean Fe concentration of 0.2nmoll-1,a dissolved Fe inventory of 0.6x109mol.the surface water residence time then calculates as 0.07-0.15years or only 1-2 months.this would be even shorter when Fe from sedimentary sources is included.simply exporting all this Fe towards the seafloor,and again assuming some 50% mobilisation from deep sea sediments,the estimated input of dissolved Fe into deep waters becomes 2.0-4.5x109 moly-1.Given the total volume of this basin,and an assumed deep water concentration of 1nmoll-1,the Fe inventory is 100x109mol and the estimated residence time is 22-50years.

Answer 2

4.4.2居留时间在个人海洋盆地

由于地理分布的非常的不均匀的湿沉积溶解的铁、非常分布不均的出口生产及其随后从海底表层沉积物动员,没有太多的效度概念,在全球的停留时间为一个元素如铁。此外,网河输入,虽然比起其他的来源谦虚,也是主要的输入分布不均中央大西洋。

因此,一个更有效的方法是构建物质平衡为每一个海洋盆地。例如,北部和赤道大西洋到50 ~ 10秒第三或3-8x109moly-1 N接待一个dissoved及其铁从输入。加入到这一个一半的全球河流输入的铁、或1.3 x109 moly-1,合并后的输入-9.3 - x109加起来moll-1。上层100米深的交易量有3 x1015m3,平均浓度为0.2 nmoll-1铁,铁库存0.6 x109mol解散。水面停留时间为0.07岁并计算1 - 2个月或者只。这将是更短的当铁从砂体的沉积来源是包括在内。这一切只是出口对海底铁,并再次假设有50%来自深海沉积物动员,估计的输入,溶解的铁抛在深坑里成为2.0 - -4.5 x109 moly-1。给出该盆地的总量,一个假定的深1 nmoll-1水浓度,将铁库存是100 x109mol和估计的停留时间22-50years . .

Answer 3

个人大洋盆地内的停留时间
由于湿沉降，溶解铁和出口生产和随后表层沉积物动员的分布很不平衡的地理分布很不平衡，有没有一个全球性的停留时间概念多有效性元素，如Fe.Moreover河流，净输入，虽然幅度不大，与其他来源相比，也分布很不均匀主要投入到大西洋中部。
因此，更有效的方法，将每个海洋basin.for例如兴建的质量平衡，北部和赤道大西洋海洋〜10秒至50 N收到从风沙三分之一或38x109moly- 1 dissoved铁input.adding这一半，铁，或1.3x109钼-1全球河流输入相结合的投入加起来有3x1015m3量4.3-9.3x109莫尔- 1，上百米的深度，并用铁的平均浓度为0.2 nmoll- 1，溶解铁库存0.6x109mol.the地表水停留时间，然后计算为0.070.15年或只有1-2 months.this会更短铁沉积来源是included.simply出口这一切铁朝海底，并再次假设一些从深海沉积物中的50％动员，估计输入到深海溶解铁变成2.0-4.5x109钼1.Given在这个盆地的总量，并假设深层水浓度1nmoll 1，铁库存100x109mol和估计的停留时间是2250年。