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\(n_{NaOH}=0,5.2=1\left(mol\right)\)
PT: \(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)
Theo PT: \(n_{HNO_3}=n_{NaNO_3}=n_{NaOH}=1\left(mol\right)\)
a, \(C_{M_{HNO_3}}=\dfrac{1}{0,3}=\dfrac{10}{3}\left(M\right)\)
b, \(C_{M_{NaNO_3}}=\dfrac{1}{0,5+0,3}=1,25\left(M\right)\)
\(n_{NaOH}=0,5.2=1\left(mol\right)\\ PTHH:NaOH+HNO_3\rightarrow NaNO_3+H_2O\\ a,n_{HNO_3}=n_{NaOH}=1\left(mol\right)\\ C_{MddHNO_3}=\dfrac{1}{0,3}=\dfrac{10}{3}\left(M\right)\\ b,V_{ddsau}=0,5+0,3=0,8\left(l\right)\\ n_{NaNO_3}=n_{NaOH}=1\left(mol\right)\\ C_{MddNaNO_3}=\dfrac{1}{0,8}=1,25\left(M\right)\)
\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)
\(n_{H_2SO_4}=0.3\cdot1=0.3\left(mol\right)\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(0.1........0.1...........0.1.......0.1\)
\(\Rightarrow H_2SO_4dư\)
\(n_{H_2SO_4\left(dư\right)}=0.3-0.1=0.2\left(mol\right)\)
\(n_{ZnSO_4}=n_{H_2}=0.1\left(mol\right)\)
\(C_{M_{ZnSO_4}}=\dfrac{0.1}{0.3}=0.33\left(M\right)\)
\(C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0.2}{0.3}=0.66\left(M\right)\)
\(a) Zn + H_2SO_4 \to ZnSO_4 + H_2\\ b) n_{Zn} = \dfrac{6,5}{65} = 0,1 < n_{H_2SO_4} =0,3 \to H_2SO_4\ dư\\ n_{H_2SO_4\ pư} = n_{ZnSO_4} = n_{Zn} = 0,1(mol)\\ n_{H_2SO_4\ dư} = 0,3 - 0,1 = 0,2(mol)\\ c) C_{M_{ZnSO_4}} = \dfrac{0,1}{0,3} = 0,33M\\ C_{M_{H_2SO_4}} = \dfrac{0,2}{0,3} = 0,67M\)
\(n_{NaOH}=0,5.0,2=0,1\left(mol\right);n_{HCl}=0,5.0,3=0,15\left(mol\right)\)
PTHH: NaOH + HCl → NaCl + H2O
Mol: 0,1 0,1 0,1
Ta có: \(\dfrac{0,1}{1}< \dfrac{0,15}{1}\) ⇒ NaOH hết, HCl dư
Vdd sau pứ = 0,2 + 0,3 = 0,5 (l)
\(C_{M_{ddNaCl}}=\dfrac{0,1}{0,5}=0,2M\)
\(C_{M_{ddHCldư}}=\dfrac{0,15-0,1}{0,5}=0,1M\)
a/ta có n(1)NaCl=0,2mol. ADCT:n=CM*Vdd. n(2)NaCl=1*0,2=0,2mol
Tổng mol của Na(1)và Na(2):
0,2+0,2=0,4mol
Vdd=0,2:1=0,2
---->CM=0,4/0,2=2(M)
b/
300mlKCl 0,4 C-1,2
CM
200mKCl 1,2. 0,4-C
300/200=C-1,2/0,4-C
-->CM=0,72(M)
Câu 1:
a) \(C\%=\dfrac{15}{15+45}.100\%=25\%\)
b) \(C_M=\dfrac{0,5}{1,5}=0,33M\)
Câu 2:
a) \(n_{NaOH}=0,5.1=0,5\left(mol\right)=>m_{NaOH}=0,5.40=20\left(g\right)\)
b) \(n_{HCl}=0,2.0,5=0,1\left(mol\right)=>m_{HCl}=0,1.36,5=3,65\left(g\right)\)
\(a,n_{NaCl}=\dfrac{5,85}{58,5}=0,1\left(mol\right)\\ C_{M\left(NaCl\right)}=\dfrac{0,1}{0,3}=\dfrac{1}{3}M\\ b,n_{CuSO_4}=\dfrac{300}{160}=1,875\left(mol\right)\\ C_{M\left(CuSO_4\right)}=\dfrac{1,875}{0,3}=6,25M\)