nhom ki hieu sao
mot loai quang boxit chua 85% nhom oxit.de dieu che duoc 1,5 tan nhom can bao nhieu tan quang biet hieu suat cua pu chi dat 60%
\(n_{Al}=\dfrac{1,5.10^6}{27}=\dfrac{1}{18}.10^6\left(mol\right)\)
\(4Al+3O_2\rightarrow2Al_2O_3\)
\(\rightarrow n_{Al_2O_3}=\dfrac{1}{36}.10^6\left(mol\right)\) < lý thuyết >
\(\rightarrow n_{Al_2O_3}=\dfrac{1}{36}.10^6:60\%=\dfrac{5}{108}.10^6\left(mol\right)\)
\(\rightarrow m_{boxit}=\dfrac{5}{108}.10^6:85\%=\dfrac{25}{459}.10^6\left(g\right)\)\(\approx54466\) (tấn)
PTHH: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Ta có: \(n_{Al}=\dfrac{1500}{27}=\dfrac{500}{9}\left(kmol\right)\) \(\Rightarrow n_{Al_2O_3}=\dfrac{250}{9}\left(kmol\right)\)
\(\Rightarrow n_{Al_2O_3\left(lýthuyết\right)}=\dfrac{\dfrac{250}{9}}{60\%}=\dfrac{1250}{27}\left(kmol\right)\) \(\Rightarrow m_{Al_2O_3\left(lýthuyết\right)}=\dfrac{1250}{27}\cdot102=\dfrac{42500}{9}\left(kg\right)\)
\(\Rightarrow m_{quặng}=\dfrac{\dfrac{42500}{9}}{85\%}=\dfrac{50000}{9}\left(kg\right)\approx5,56\left(tấn\right)\)
nguyen to a co so hieu nguyen tu la 11 , chu ki 3,nhom I trong bang tuan hoan cac nguyen to hoa hoc.hay cho biet:
-cau tao nguyen tu cua A.
-tinh chat hoa hoc dac trung cua a
-so sanh tinh chat hoa hoc cua a voi nguyen to lan can
https://olm.vn/hoi-dap/detail/240862214307.html và https://d3.violet.vn//uploads/previews/present/4/447/16/preview.swf
Bạn băng băng không giỏi gì đâu ,copy mạng ý mà
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