如何将摩尔总质量(e / m ^ 2。g克,公斤等)?- 江南体育网页版- - - - -地球科学堆江南电子竞技平台栈交换 最近30从www.hoelymoley.com 2023 - 03 - 31 - t11:59:45z //www.hoelymoley.com/feeds/question/19444 https://creativecommons.org/licenses/by-sa/4.0/rdf //www.hoelymoley.com/q/19444 6 如何将摩尔总质量(e / m ^ 2。g克,公斤等)? Niyamat Ullah //www.hoelymoley.com/users/19125 2020 - 03 - 12 - t14:39:24z 2020 - 03 - 12 - t18:17:38z < p >我要计算总类< span = " math-container " > $ \ rm {NO_2} $ < / span >数量在一年内使用卫星< a href = " https://developers.google.com/earth-engine/datasets/catalog/COPERNICUS_S5P_OFFL_L3_NO2 " rel = " nofollow noreferrer " > < / > Sentinel-5p NO2数据集。但问题是哨兵卫星数据存储在<跨类= " math-container " > \ rm \美元压裂{摩尔}{m ^ 2} $ < / span >单位。但是我必须把它比作<代码>磅< /代码>或<代码>吨> < /代码或任何其他单位比较<跨类= " math-container " > $ \ rm {NO_2} $ < / span >数量与地面<跨类= " math-container " > $ \ rm {NO_2} $ < / span >测量。< / p > < blockquote > < p > <强>例子:< /强>在我们的卫星数据集,我们<强>对流层<跨类= " math-container " > $ \ rm {NO_2} $ < / span >列数密度< / >强的单位是美元<跨类=“math-container”> \ rm \压裂{摩尔}{m ^ 2} $ < / span >。我们要计算总年度<跨类= " math-container " > $ \ rm {NO_2} $ < / span >加州。今年我们有总结所有加州的数据。然后我们得到一个总年度<跨类= " math-container " > $ \ rm {NO_2} $ < / span > 4200 <跨类= " math-container " > \ rm \压裂{摩尔}{m ^ 2} $ < / span >。我想把这<代码>克< /代码>或<代码>吨< /代码>或<代码>磅> < /代码。例如,每年总类< span = " math-container " > $ \ rm {NO_2} $ < / span >加州是35000吨。< / p >

Also, if I convert that $\rm\frac{mol}{m^2}$ to $\rm\frac{molec}{cm^2}$, is it then possible to convert it to total gram or ton?

Research: I have asked that question in here. After getting answers, I thought I have found the solutions. But after some more research, it turns out not easy. Because in satellite data what we get is tropospheric vertical column number density and convert them to total mass like total $\rm{NO_2}$ in gram or ton is not straightforward. These forum posts (#1, #2, #3) also somewhat similar to mine, but I didn't get quite a solution to my problem.

EDIT: I know that we can easily multiply by area in $\rm m^2$ to cancel out $\rm m^2$ in the unit and multiple by $\rm{NO_2}$ molar mass, but this is not a ground-based problem. We are talking about satellite imagery dataset especially tropospheric vertical column density. #1, #2, #3 These posts shows why it is different for converting satellite imagery units to ground-based units.

//www.hoelymoley.com/questions/19444/how-to-convert-mol-m2-to-total-mass-e-g-gram-kg-etc/19447 # 19447 7 答案为如何将摩尔peterh总质量(e / m ^ 2。g克,公斤等)? peterh //www.hoelymoley.com/users/6016 2020 - 03 - 12 - t15:33:36z 2020 - 03 - 12 - t15:33:36z < p > < span class = " math-container”> \ rm \美元压裂{摩尔}{m ^ 2} $ < / span >显示的数量< span class = " math-container " > $ \ rm {NO_2} $ < / span >大气中在一平方米的面积——摩尔。< / p > < p >的摩尔质量< span class = " math-container " > $ \ rm {NO_2} $ < / span >是<跨类= " math-container " > 14美元+ 2 \ cdot 16 = 46 $ < / span >。这意味着,1摩尔质量的< span class = " math-container " > $ \ rm {NO_2} $ < / span >是<跨类= " math-container " > $ \ rm {46 g} $ < / span >。< / p > < p >地球表面积是5.1亿<跨类= " math-container " > \ rm{公里^ 2}$ < / span >。因此,1 < span class = " math-container”> \ rm \美元压裂{摩尔}{m ^ 2} \ rm {NO_2} < / span >美元转换成< span class = " math-container " > \美元rm{46 \压裂{g} {m ^ 2} \ cdot 5.1 \ cdot 10 ^ 8公里^ 2}< / span >美元,意味着总质量的< span class = " math-container " > 23.46美元\ cdot 10 ^ 9 \ rm台币< / span >在整个气氛。< / p > < p >这里我们假定一个常数分布的< span class = " math-container " > $ \ rm {NO_2} $ < / span >在大气中,这通常不是在地球科学相关的问题。江南体育网页版还你的研究问题是高度非水平分布的可视化。在这里你需要集成(或者,考虑你的数字环境,总和)。< / p >
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