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a, \(n_{NaOH}=0,2.1=0,2\left(mol\right)\)
\(m_{NaOH}=0,2.40=8\left(g\right)\)
b, \(n_{H_2SO_4}=2.0,1=0,2\left(mol\right)\)
\(c,C\%=\dfrac{6}{200}.100\%=3\%\)
\(m_{NaCl}=\dfrac{200.8}{100}=16\left(g\right)\)
a) \(m_{HCl}=200\cdot7,3\%=14,6\left(g\right)\)
b) \(n_{NaOH}=0,5\cdot1=0,5\left(mol\right)\) \(\Rightarrow m_{NaOH}=0,5\cdot40=20\left(g\right)\)
c) \(n_{CuSO_4}=0,2\cdot1,5=0,3\left(mol\right)\) \(\Rightarrow m_{CuSO_4}=0,3\cdot160=48\left(g\right)\)
d) Bạn xem lại đề !
a) mHCl=200⋅7,3%=14,6(g)mHCl=200⋅7,3%=14,6(g)
b) nNaOH=0,5⋅1=0,5(mol)nNaOH=0,5⋅1=0,5(mol) ⇒mNaOH=0,5⋅40=20(g)⇒mNaOH=0,5⋅40=20(g)
c) nCuSO4=0,2⋅1,5=0,3(mol)nCuSO4=0,2⋅1,5=0,3(mol) ⇒mCuSO4=0,3⋅160=48(g)⇒mCuSO4=0,3⋅160=48(g)
d) Bạn xem lại đề !
\(n_{NaOH}=0.2\cdot0.5=0.1\left(mol\right)\)
\(n_{CuSO_4}=0.1\cdot2=0.2\left(mol\right)\)
\(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\)
\(0.1.............0.05...............0.05...........0.05\)
\(m_{Cu\left(OH\right)_2}=0.05\cdot98=4.9\left(g\right)\)
\(C_{M_{Na_2SO_4}}=\dfrac{0.05}{0.2+0.1}=0.167\left(M\right)\)
\(C_{M_{CuSO_4\left(dư\right)}}=\dfrac{0.2-0.05}{0.1}=1.5\left(M\right)\)
\(a,C\%_{NaCl}=\dfrac{15}{15+185}.100\%=7,5\%\\ b,m_{HNO_3}=\dfrac{18,9}{100}.100+\dfrac{6,3}{100}.200=31,5\left(g\right)\\ m_{ddHNO_3}=100+200=300\left(g\right)\\ C\%_{HNO_3}=\dfrac{31,5}{300}.100\%=10,5\%\)
\(c,n_{NaCl}=\dfrac{5,85}{58,5}=0,1\left(mol\right)\\ C_{M\left(NaCl\right)}=\dfrac{0,1}{0,1}=1M\\ d,n_{KOH}=2.0,2+0,2.0,2=0,44\left(mol\right)\\ V_{ddKOH}=0,2+0,2=0,4\left(l\right)\\ C_{M\left(KOH\right)}=\dfrac{0,44}{0,4}=1,1M\\ e,m_{NaOH}=\dfrac{150.16}{100}=24\left(g\right)\\ m_{ddNaOH}=50+150=200\left(g\right)\\ C\%_{NaOH}=\dfrac{24}{200}.100\%=12\%\)
a)
\(m_{H_2SO_4}=\dfrac{300.19,6}{100}=58,8\left(g\right)\)
=> \(m_{dd.H_2SO_4.9,8\%}=\dfrac{58,8.100}{9,8}=600\left(g\right)\)
=> \(m_{H_2O\left(thêm\right)}=600-300=300\left(g\right)\)
b)
\(n_{HCl}=0,2.2=0,4\left(mol\right)\)
=> \(V_{dd.HCl.1,5M}=\dfrac{0,4}{1,5}=\dfrac{4}{15}\left(l\right)\)
=> \(V_{H_2O\left(thêm\right)}=\dfrac{4}{15}-0,2=\dfrac{1}{15}\left(l\right)=\dfrac{200}{3}\left(ml\right)\)
=> \(m_{H_2O\left(thêm\right)}=\dfrac{200}{3}.1=\dfrac{200}{3}\left(g\right)\)