Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A/
Cu(OH)2 => (to) CuO + H2O
CuO + H2 => (to) Cu + H2O
Cu +Cl2 => CuCl2
A/
Cu(OH)2\(\xrightarrow[]{to}\) CuO+ H2O
CuO+ H2\(\xrightarrow[]{to}\) Cu+ H2O
Cu+ 2FeCl3\(\rightarrow\) 2FeCl2+ CuCl2
B/
CuO+ CO\(\xrightarrow[]{to}\) Cu+ CO2
Cu+ \(\frac{1}{2}\)O2+ H2O\(\rightarrow\) Cu(OH)2
Cu(OH)2+ 2HCl\(\rightarrow\) CuCl2+ 2H2O
C/
CuO+ C2H5OH\(\rightarrow\) CH3CHO+ Cu(OH)2\(\downarrow\)+ H2O
Cu(OH)2+ 2HCl\(\rightarrow\) CuCl2+ 2H2O
CuCl2+ Fe\(\rightarrow\) FeCl2+ Cu
D/
2Cu+ O2\(\xrightarrow[]{to}\) 2CuO
CuO+ C2H5OH\(\rightarrow\) CH3CHO+ Cu(OH)2\(\downarrow\)+ H2O
Cu(OH)2+ 2HCl\(\rightarrow\) CuCl2+ 2H2O
\(Cu\overset{\left(1\right)}{-->}CuO\overset{\left(2\right)}{-->}CuSO_4\overset{\left(3\right)}{-->}CuCl_2\overset{\left(4\right)}{-->}Cu\left(NO_3\right)_2\overset{\left(5\right)}{-->}Cu\left(OH\right)_2\overset{\left(6\right)}{-->}CuO\overset{\left(7\right)}{-->}Cu\)
\(\left(1\right)2Cu+O_2\overset{t^o}{--->}2CuO\)
\(\left(2\right)CuO+H_2SO_4--->CuSO_4+H_2O\)
\(\left(3\right)CuSO_4+BaCl_2--->BaSO_4\downarrow+CuCl_2\)
\(\left(4\right)CuCl_2+2AgNO_3--->2AgCl\downarrow+Cu\left(NO_3\right)_2\)
\(\left(5\right)Cu\left(NO_3\right)_2+2NaOH--->Cu\left(OH\right)_2\downarrow+2NaNO_3\)
\(\left(6\right)Cu\left(OH\right)_2\overset{t^o}{--->}CuO+H_2O\)
\(\left(7\right)CuO+H_2\overset{t^o}{--->}Cu+H_2O\)
\(2Cu+O_2\underrightarrow{t^o}2Cuo\)
\(CuO+H_2SO_4->CuSO_4+H_2O\)
\(CuSO_4+BaCl_2->BaSO_4\downarrow+CuCl_2\)
\(CuCl_2+2AgNO_3->Cu\left(NO_3\right)_2+2AgCl\downarrow\)
\(Cu\left(NO_3\right)_2+2NaOH->Cu\left(OH\right)_2\downarrow+2NaNO_3\)
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
a)
(1) \(Cu+2H2SO4\left(đ\right)-^{t0}->CuSO4+SO2\uparrow+2H2O\)
\(\left(2\right)CuSO4+BaCl2->CuCl2+BaSO4\downarrow\)
(3) \(CuCl2+2AgNO3->Cu\left(NO3\right)2+2AgCl\downarrow\)
(4) \(Cu\left(NO3\right)2+2NaOH->Cu\left(OH\right)2\downarrow+2NaNO3\)
(5) \(Cu\left(OH\right)2-^{t0}->CuO+H2O\)
(6) \(CuO+2HCl->CuCl2+H2O\)
(7) \(CuCl2-^{t0}->Cu+Cl2\uparrow\)
(8) \(Cu+Cl2-^{t0}->CuCl2\)
(9) \(CuCl2+2NaOh->Cu\left(OH\right)2\downarrow+2NaCl\)
b)
\(4Na+O2-^{t0}->2Na2O\)
\(Na2O+H2O->2NaOH\)
\(2NaOH+CO2->Na2CO3+H2O\)
\(Na2CO3+H2SO4->Na2SO4+CO2\uparrow+H2O\)
\(Na2SO4+BaCl2->2NaCl+BaSO4\downarrow\)
\(2NaCl+2H2O-\xrightarrow[có-màng-ngăn]{đpdd}2NaOH+Cl2\uparrow+H2\uparrow\)
\(3NaOH+H3PO4->Na3PO4+3H2O\)
Tham khảo
(1) 2Cu + O2 ---> 2CuO
(2) CuO + H2SO4 ---> CuSO4 + H2
(3) CuSO4 + 2NaOH ---> Cu(OH)2 + Na2SO4
(4) Cu(OH)2 + 2HCl ---> CuCl2 + 2H2O
(5) CuCl2 + 2AgNO3 ----> Cu(NO3)2 + 2AgCl
\(2Cu+O_2\underrightarrow{t^o}2CuO\)
\(CuO+2HCl->CuCl_2+H_2O\)
\(CuCl_2+2NaOH->Cu\left(OH\right)_2\downarrow+2NaCl\)
\(Cu\left(OH\right)_2+H_2SO_4->CuSO_4+2H_2O\)
\(CuSO_4+Ba\left(NO_3\right)_2->Cu\left(NO_3\right)_2+BaSO_4\downarrow\)
\(\left(1\right)CuO+2HCl\rightarrow CuCl_2+H_2O\\ \left(2\right)CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2\downarrow+KCl\\ \left(3\right)Cu\left(OH\right)_2+H_2SO_4\rightarrow CuSO_4+2H_2O\\ \left(4\right)CuSO_4+BaCl_2\rightarrow CuCl_2+BaSO_4\downarrow\)
$Cu(OH)_2 \xrightarrow{t^o} CuO + H_2O$
$CuO + 2HCl \to CuCl_2 + H_2o$
$CuCl_2 + 2AgNO_3 \to Cu(NO_3)_2 + 2AgCl$
$Cu(NO_3)_2 + 2KOH \to Cu(OH)_2 + 2KNO_3$
giúp e câu này với ah
Câu 11: Nhiệt phân hoàn toàn 30,3 gam hỗn hợp 2 bazơ không tan Cu(OH)2 và Fe(OH)3, sau pư thu được 24 gam 2 oxit.
a) Tính % số mol mỗi bazơ trong hỗn hợp đầu?
Cho 2 oxit trên pư vừa đủ với 200 ml dung dịch HCl xM. Tìm x?
\(\left(1\right)Cu+Cl_2\overset{t^o}{--->}CuCl_2\)
\(\left(2\right)CuCl_2+2NaOH--->Cu\left(OH\right)_2+2NaCl\)
\(\left(3\right)Cu\left(OH\right)_2\overset{t^o}{--->}CuO+H_2O\)
\(\left(4\right)CuO+2HCl--->CuCl_2+H_2O\)
Cu+Cl2-to>CuCl2
CuCl2+2NaOH->2NaCl+Cu(OH)2
Cu(oH)2->CuO+H2O
CuO+2HCl->CuCl2+H2O