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\(a.V_{CO_2\left(dktc\right)}=0,25.22,4=5,6\left(l\right)\)
\(b.m_{Al_2O_3}=0,5.160=80\left(g\right)\)
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)
Câu 1: B
\(n_{H_2S}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\) => \(m_{H_2S}=0,4.34=13,6\left(g\right)\)
Câu 2: C
\(m_{CO_2}=1.44=44\left(g\right);n_{N_2O}=1.44=44\left(g\right)\)
Câu 3: B
2Fe(OH)3 + 3H2SO4 --> Fe2(SO4)3 + 6H2O
Câu 4: C
CTHH: CaSO4
PTK = 40.1 + 32.1 + 16.4 = 136 (đvC)
Câu 5: C
\(M_X=4,5.24=108\left(đvC\right)\)
=> X là Ag
\(a.n_{NaOH}=\dfrac{0,4}{40}=0,01\left(mol\right)\\ b.n_{H_2O}=\dfrac{0,6.10^{23}}{6.10^{23}}=0,1\left(mol\right)\\ m_{H_2O}=0,1.18=1,8\left(g\right)\)
\(c.n_{O_2}=\dfrac{9,6}{16}=0,6\left(mol\right)\\ V_{O_2}=0,6.22,4=13,44\left(l\right)\\ d.n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ Số.phân.tử.là:0,25.6.10^{23}=1,5.10^{23}\left(phân.tử\right)\)
a.
\(m_{Al}=0.5\cdot27=13.5\left(g\right)\)
\(m_{CO_2}=\dfrac{6.72}{22.4}\cdot44=13.2\left(g\right)\)
\(m_{N_2}=\dfrac{5.6}{22.4}\cdot28=7\left(g\right)\)
\(m_{CaCO_3}=0.25\cdot100=25\left(g\right)\)
b.
\(m_{hh}=\dfrac{3.36}{22.4}\cdot2+\dfrac{5.6}{22.4}\cdot28+0.2\cdot44=16.1\left(g\right)\)
Câu 1 :
a)
nCO2 = 13.2/44 = 0.3(mol)
VCO2 = 0.3*22.4 = 6.72 (l)
b)
nC4H10 = 8.96/22.4 = 0.4 (mol)
mC4H10 = 0.4*58 = 23.2 (g)
c)
nCaO = 3*10^23 / 6 *10^23 = 0.5 (mol)
nCa(OH)2 = 1.8*10^23 / 6*10^23 = 0.3 (mol)
mA = 0.5*56 + 0.3*74 = 50.2 (g)
Câu 1::
a) nCO2=13,2/44=0,3(mol)
=>V(CO2,đktc)=0,3.22,4=6,72(l)
b) nC4H10=8,96/22,4=0,4(mol)
->mC4H10=0,4.58= 23,2(g)
c) nCaO= (3.1023)/(6.1023)= 0,5(mol)
nCa(OH)2= (1,8.1023)/(6.1023)=0,3(mol)
=>mhhA= mCaO+ mCa(OH)2= 0,5.56 + 0,3.74= 50,2(g)