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\(a_1,m_{CaCO_3}=0,25.100=25(g)\\ a_2,m_{SO_2}=\dfrac{3,36}{22,4}.64=9,6(g)\\ a_3,m_{H_2SO_4}=\dfrac{9.10^{23}}{6.10^{23}}.98=147(g)\)
a) nNaOH=20/40=0,5(mol)
nN2=1,12/22,4=0,05(mol)
nNH3= (0,6.1023)/(6.1023)=0,1(mol)
b) mAl2O3= 102.0,15= 15,3(g)
mSO2= nSO2 . M(SO2)= V(CO2,đktc)/22,4 . 64= 6,72/22,4. 64= 0,3. 64= 19,2(g)
mH2S= nH2S. M(H2S)= (0,6.1023)/(6.1023) . 34=0,1. 34 = 3,4(g)
c) V(CO2,đktc)=0,2.22.4=4,48(l)
nSO2=16/64=0,25(mol) -> V(SO2,đktc)=0,25.22,4=5,6(l)
nCH4=(2,1.1023)/(6.1023)=0,35(mol) -> V(CH4,đktc)=0,35.22,4=7,84(l)
a) mO2= nO2. M(O2)=0,45. 32=14,4(g)
b) mBaCO3=nBaCO3.M(BaCO3)=0,6.197=118,2(g)
c) mAl2(SO4)3=nAl2(SO4)3.M(Al2(SO4)3)=1,5.342=513(g)
d) nSO2=V(SO2,đktc)/22,4=16,8/22,4=0,7(mol)
=> mSO2=nSO2.M(SO2)=0,7.64=44,8(g)
e) nH2O=(3.1023):(6.1023)=0,5(mol)
=>mH2O=nH2O.M(H2O)=0,5.18=9(g)
f) nCO2=V(CO2,đktc)/22,4=8,96/22,4=0,4(mol)
=>mCO2=nCO2.M(CO2)=0,4.44=17,6(g)
\(a,n_{CO_2}=\dfrac{m_{CO_2}}{M_{CO_2}}=\dfrac{11}{44}=0,25\left(mol\right)\\ b,n_{H_2}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\\ V_{H_2\left(đktc\right)}=1,5.22,4=33,6\left(l\right)\)
a) nCO2=[(9.1023)/(6.1023)]=1,5(mol)
=> mCO2=1,5.44=66(g)
V(CO2,đktc)=1,5.22,4=33,6(l)
b) nH2=4/2=2(mol)
N(H2)=2.6.1023=12.1023(phân tử)
V(H2,đktc)=2.22,4=44,8(l)
c) N(CO2)=0,5.6.1023=3.1023(phân tử)
V(CO2,đktc)=0,5.22,4=11,2(l)
mCO2=0,5.44=22(g)
d) nN2=2,24/22,4=0,1(mol)
mN2=0,1.28=2,8(g)
N(N2)=0,1.1023.6=6.1022 (phân tử)
e) nCu=[(3,01.1023)/(6,02.1023)]=0,5(mol)
mCu=0,5.64=32(g)
Mà sao tính thể tích ta :3
Bài 6:
a) \(m_{CuSO_4}=n.M=0,6.160=96\left(g\right)\)
b) \(n_{Cl_2}=\dfrac{\text{Số phân tử}}{6.10^{^{23}}}=\dfrac{3.10^{^{23}}}{6.10^{^{23}}}=0,5\left(mol\right)\)
`=>` \(m_{Cl_2}=n.M=0,5.71=35,5\left(g\right)\)
c) \(n_{CH_4}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
=>`` \(m_{CH_4}=n.M=0,5.16=8\left(g\right)\)
d) \(m_{C_{12}H_{22}O_{11}}=1,5.342=513\left(g\right)\)
Bài 7:
a) \(n_{CO_2}=\dfrac{m}{M}=\dfrac{11}{44}=0,25\left(mol\right)\)
`=>` \(V_{CO_2\left(đktc\right)}=n.22,4=0,25.22,4=5,6\left(l\right)\)
b) \(n_{H_2S}=\dfrac{\text{Số phân tử}}{6.10^{23}}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)
`=>` \(V_{H_2S\left(đktc\right)}=n.22,4=1,5.22,4=33,6\left(l\right)\)
c) \(V_{Cl_2\left(đktc\right)}=n.22,4=0,7.22,4=15,68\left(l\right)\)
d) N phân tử H2 có 6.1023 phân tử H2
`=>` \(n_{H_2}=\dfrac{\text{Số phân tử}}{6.10^{23}}=\dfrac{6.10^{23}}{6.10^{23}}=1\left(mol\right)\)
`=>` \(V_{H_2\left(đktc\right)}=n.22,4=1.22,4=22,4\left(l\right)\)
\(6a)m_{CuSO_4}=n.M=0,6.160=96\left(g\right)\)
\(b)n_{Cl_2}=\dfrac{\text{Số phân tử }Cl_2}{N}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)
\(m_{Cl_2}=n.M=0,5.71=35,5\left(g\right)\)
\(c)n_{CH_4}=\dfrac{V}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(m_{CH_4}=n.M=0,5.16=8\left(g\right)\)
\(d)m_{C_6H_{12}O_6}=n.M=1,5.180=270\left(g\right)\)
\(7a)n_{CO_2}=\dfrac{m}{M}=\dfrac{11}{44}=0,25\left(mol\right)\)
\(V_{CO_2}=n.22,4=0,25.22,4=5,6\left(l\right)\)
\(b)n_{H_2S}=\dfrac{\text{Số phân tử }H_2S}{N}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)
\(V_{H_2S}=n.22,4=1,5.22,4=33,6\left(l\right)\)
\(c)V_{Cl_2}=n.22,4=0,7.22,4=15,68\left(l\right)\)
\(d)n_{H_2}=\dfrac{\text{Số phân tử }H_2}{N}=\dfrac{6.10^{23}}{6.10^{23}}=1\left(mol\right)\)
\(V_{H_2}=n.22,4=1.22,4=22,4\left(l\right)\)