25,65+12,56=?
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
12,56 x 127 - 12,56 x 27 + 0,8744 : 100
= 12,56 x ( 127 - 27 ) + 0,008744
= 12,56 x 100 + 0,008744
= 1256 + 0,008744
= 1256,008744
\(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ n_{Ba\left(OH\right)_2}=\dfrac{200.25,65\%}{171}=0,3\left(mol\right)\\ Vì:0,5< \dfrac{n_{Ba\left(OH\right)_2}}{n_{CO_2}}=\dfrac{0,3}{0,4}=0,75< 1\\ \Rightarrow Sp:\left\{{}\begin{matrix}BaCO_3:a\left(mol\right)\\Ba\left(HCO_3\right)_2:b\left(mol\right)\end{matrix}\right.\\\Rightarrow \left\{{}\begin{matrix}a+b=0,3\\a+2b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ \Rightarrow m_{\downarrow}=m_{BaCO_3}=197.0,2=39,4\left(g\right)\)
\(34,65+12,56-4,65-15,46?\)
\(=\left(34,65-4,65\right)-\left(15,46-12,56\right)\)
\(=30-2,9\)
\(=27,1\)
em cảm ơn anh chị nhìu, oh mà khoan, anh hay là chị ?????
\(n_{SO_2}=\dfrac{3,2}{64}=0,05\left(mol\right)\\ m_{Ba\left(OH\right)_2}=\dfrac{200\cdot25,65\%}{100\%}=51,3\left(g\right)\\ \Rightarrow n_{Ba\left(OH\right)_2}=\dfrac{51,3}{171}=0,3\left(mol\right)\\ PTHH:SO_2+Ba\left(OH\right)_2\rightarrow BaSO_3\downarrow+H_2O\\ \text{Vì }\dfrac{n_{SO_2}}{1}< \dfrac{n_{Ba\left(OH\right)_2}}{1}\text{ nên sau p/ứ }Ba\left(OH\right)_2\text{ dư}\\ \Rightarrow n_{BaSO_3}=n_{H_2O}=0,05\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}m_{CT_{BaSO_3}}=0,05\cdot217=10,85\left(g\right)\\m_{H_2O}=0,05\cdot18=0,9\left(g\right)\end{matrix}\right.\\ \Rightarrow m_{dd_{BaSO_3}}=3,2+200-0,9=202,3\left(g\right)\\ \Rightarrow C\%_{BaSO_3}=\dfrac{10,85}{202,3}\cdot100\%\approx5,36\%\)
25,65 + 12,56 = 38,21
25,65+12,56=38,21