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![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Ag_2O}=\dfrac{4.64}{232}=0.02\left(mol\right)\)
\(m_{dd_{HNO_3}}=300\cdot1.59=477\left(g\right)\)
\(Ag_2O+2HNO_3\rightarrow2AgNO_3+H_2O\)
\(0.02............................0.04\)
\(m_{AgNO_3}=0.04\cdot170=6.8\left(g\right)\)
\(m_{dd}=4.64+477=481.64\left(g\right)\)
\(C\%_{AgNO_3}=\dfrac{6.8}{481.64}\cdot100\%=1.4\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
300ml = 0,3l
\(n_{HNO3}=1.0,3=0,3\left(mol\right)\)
Pt : \(NaOH+HNO_3\rightarrow NaNO_3+H_2O|\)
1 1 1 1
0,3 0,3 0,3
\(n_{NaOH}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
200ml = 0,2l
\(C_{M_{NaOH}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)
\(n_{NaNO3}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
⇒ \(m_{NaNO3}=0,3.85=25,5\left(g\right)\)
Sau phản ứng :
\(V_{dd}=0,2+0,3=0,5\left(l\right)\)
\(C_{M_{NaNO3}}=\dfrac{0,3}{0,5}=0,6\left(M\right)\)
Chúc bạn học tốt
\(n_{HNO_3}=0,3\left(mol\right)\)
\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)
Theo PT: \(n_{NaOH}=n_{NaNO_3}=n_{HNO_3}=0,3\left(mol\right)\)
\(\Rightarrow CM_{NaOH}=\dfrac{0,3}{0,2}=1,5M\)
\(m_{NaNO_3}=0,3.85=25,5\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi \(\left\{{}\begin{matrix}C_{M\left(HNO_3\right)}=aM\\C_{M\left(KOH\right)}=bM\end{matrix}\right.\)
Phần 1:
nHNO3 = 0,03a (mol)
nKOH = 0,09b (mol)
PTHH: KOH + HNO3 --> KNO3 + H2O
=> 0,03a = 0,09b
=> a = 3b
Phần 2:
nHNO3 = 0,04a (mol)
nKOH = 0,02b (mol)
\(n_{MgO}=\dfrac{0,6}{40}=0,015\left(mol\right)\)
PTHH: MgO + 2HNO3 --> Mg(NO3)2 + H2O
____0,015-->0,03
KOH + HNO3 --> KNO3 + H2O
0,02b->0,02b
=> 0,02b + 0,03 = 0,04a
=> a = 0,9 ; b = 0,3
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\)
PTHH: Cu(OH)2 --to--> CuO + H2O
______0,3<-------------0,3
=> mCu(OH)2 = 0,3.98 = 29,4 (g)
b)
PTHH: CuO + 2HNO3 --> Cu(NO3)2 + H2O
______0,3--->0,6---------->0,3
=> \(C_{M\left(HNO_3\right)}=\dfrac{0,6}{0,2}=3M\)
mCu(NO3)2 = 0,3.188 = 56,4 (g)
\(m_{ddHNO3}=50.1,25=62,5\left(g\right)\)
b) \(m_{hno3}=\frac{62,5.40}{100}=25\left(g\right)\)
c) \(n_{HNO3}=\frac{25}{63}=0,4\left(mol\right)\)
\(C_M=\frac{0,4}{0,05}=8\left(M\right)\)