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a) mFeSO4= 0,25.152=38(g)
b) mFeSO4= \(\dfrac{13,2.10^{23}}{6.10^{23}}.152=334,4\left(g\right)\)
c) mNO2= \(\dfrac{8,96}{22,4}.46=18,4\left(g\right)\)
d) mA= 27.0,22+64.0,25=21,94(g)
e) mB= \(\dfrac{11,2}{22,4}.32+\dfrac{13,44}{22,4}.28=32,8\left(g\right)\)
g) mC= \(64.0,25+\dfrac{15.10^{23}}{6.10^{23}}.56=156\left(g\right)\)
h) mD= \(0,25.32+\dfrac{11,2}{22,4}.44+\dfrac{2,7.10^{23}}{6.10^{23}}.28=42,6\left(g\right)\)
hơi muộn nha<3
nSO2 = 20,26 / 22,4 = 0,9 (mol)
nCO = \(\frac{7,2.10^{23}}{6.10^{23}}=1,2\left(mol\right)\)
=> Khối lượng hỗn hợp là:
mhỗn hợp = 5,6 + 0,9 x 64 + 1,2 x 28 + 0,5 x 32 = 112,8 (gam)
a,Đặt tạm Al2(SO4)3 là A nhé 
Có: nA=\(\frac{m_A}{M_A}\)=\(\frac{75,24}{27.2+\left(32+16.4\right).3}\)=0,22(mol)
b,Tương tự: nA=\(\frac{V_{A\left(Đktc\right)}}{22,4}\)=0.7(mol)
n của h2=1.2.1023:6.1023=0.2 mol
nSo2=6,4:64=0.1 mol
a,Vhh=[1,5+2,5+0.2+0,1] .22,4=96,32l
mhh=(1,5.32)+(2,5.28)+(0,2.2)+6,4=124,8g
nSO2 = 6,4 / 64 = 0,1 mol
nH2 = \(\frac{1,2\times10^{23}}{6\times10^{23}}=0,2\left(mol\right)\)
a/ Vhỗn hợp khí(đktc) = ( 0,1 + 0,2 + 1,5 + 2,5 ) x 22,4 = 96,32 lít
b/ mO2 = 1,5 x 32 = 48 gam
nN2 = 2,5 x 28 = 70 gam
nH2 = 0,2 x 2 = 0,4 gam
=> mhỗn hợp khí = 48 + 70 + 0,4 + 6,4 = 124,8 gam
Câu 1 :
P(III) và O: P3O4
N(III) và H : NH3
Fe(II) và O : FeO
Cu(II) và OH : Cu(OH)2
Ca và NO3 :Ca(NO3)3
Ag và SO4 :Ag2SO4
NH4(I) và NO3 : NH4.NO3
Câu 2 :
a, \(M_{CO2}=\frac{11}{44}=0,25\left(mol\right)\)
b,
\(n_{H2}=\frac{9,10^{23}}{6.10^{23}}=1,5\left(mol\right)\)
\(\Rightarrow V_{H2}=1,5.22,4=33,6\left(l\right)\)
Câu 3 :
\(n_{H2}=\frac{1,2.10^{23}}{6.10^{23}}=0,2\left(mol\right)\)
\(n_{SO2}=\frac{6,4}{64}=0,1\left(mol\right)\)
\(V_{hh}=\left(1,5+2,5+0,2+0,1\right).22,4=96,32\left(l\right)\)
\(\Rightarrow m_{hh}=m_{O2}+m_{H2}+m_{N2}+m_{SO2}\)
\(=48+0,2+72+6,4=124,6\left(g\right)\)
Bài 5:
Ta có mS = 0,5 . 32 = 16 ( gam )
=> mFe = 56 . 0,6 = 33,6 ( gam )
=> mFe2O3 = 160 . 0,8 = 128 ( gam )
=> mhỗn hợp = 16 + 33,6 + 128 = 177,6 ( gam )
1.
Gọi CTHH của HC là A2O3
A2O3 + 3H2SO4 →→A2(SO4)3 + 3H2O
mH2SO4=12,25.24100=2,94(g)12,25.24100=2,94(g)
nH2SO4=2,9498=0,03(mol)2,9498=0,03(mol)
Theo PTHH ta có:
1313nH2SO4=nA2O3=0,01(mol)
MA2O3=1,020,01=1021,020,01=102
MA=102−16.32=27102−16.32=27
Vậy A là Al,CTHH của oxit là Al2O3
2.
a;
-nO2=2,4.10236.1023=0,4(mol)2,4.10236.1023=0,4(mol)⇒⇒VO2=22,4.0,4=8,96(lít)
-nCO2=9.10236.1023=1,5(mol)9.10236.1023=1,5(mol)⇒⇒VCO2=1,5.22,4=33,6(lít)
-nO2=6,432=0,2(mol)6,432=0,2(mol);nN2=22,428=0,8(mol)22,428=0,8(mol)⇒⇒Vhh=(0,8+0,2).22,4=22,4(lí t)(0,8+0,2).22,4=22,4(lít)
-Vhh=(0,75+0,5+0,25).22,4=33,6(lí t)(0,75+0,5+0,25).22,4=33,6(lít)
b;
-mNaOH=40.2=80(g)
-nMg=1,32.10226.1023=0,022(mol)1,32.10226.1023=0,022(mol)⇒⇒mMg=0,022.24=0,528(g)
-nHCl=9,6.10226.1023=0,16(mol)⇒9,6.10226.1023=0,16(mol)⇒mHCl=36,5.0,16=5,84(g)
-nCO2=3322,4=165112(mol)3322,4=165112(mol)
nCO=11,222,4=0,5(mol)11,222,4=0,5(mol)
nN2=5,522,4=55224(mol)5,522,4=55224(mol)
mhh=44.165112165112+28.0,5+28.5522455224=85,7(g)
\(n_{H_2}=\dfrac{1,2\times10^{23}}{6\times10^{23}}=0,2\left(mol\right)\)
\(n_{SO_2}=\dfrac{6,4}{64}=0,1\left(mol\right)\)
Ta có: \(V_{hh}=V_{O_2}+V_{N_2}+V_{H_2}+V_{SO_2}\)
\(\Leftrightarrow V_{hh}=22,4\times1,5+22,4\times2,5+22,4\times0,2+22,4\times0,1=96,32\left(l\right)\)
Ta có: \(m_{hh}=m_{O_2}+m_{N_2}+m_{H_2}+m_{SO_2}\)
\(\Leftrightarrow m_{hh}=1,5\times32+2,5\times28+0,2\times2+6,4=124,8\left(g\right)\)
\(a_1.m_{CaCl_2}=n.M=0,25.111=27,75\left(g\right)\\ m_{Cu\left(OH\right)_2}=n.M=0,5.98=49\left(g\right)\\ \Rightarrow m_{hh}=m_{CaCl_2}+m_{Cu\left(OH\right)_2}=27,75+49=76,75\left(g\right)\)
\(b_1.n_{N_2}=\dfrac{V_{\left(dktc\right)}}{22,4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ n_{O_2}=\dfrac{V_{\left(dktc\right)}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ m_{N_2}=n.M=0,25.28=7\left(g\right)\\ m_{O_2}=n.M=0,15.32=4,8\left(g\right)\\ \Rightarrow m_{hh}=m_{N_2}+m_{O_2}=7+4,8=11,8\left(g\right)\)
\(c_1.n_{H_2O}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\\ n_{CO}=\dfrac{2,4.10^{23}}{6.10^{23}}=0,4\left(mol\right)\\ m_{H_2O}=n.M=0,5.18=9\left(g\right)\\ m_{CO}=n.M=0,4.28=11,2\left(g\right)\\ m_{hh}=m_{H_2O}+m_{CO}=9+11,2=20,2\left(g\right)\)
\(a_2.n_{hh}=n_{Cl_2}+n_{N_2}=0,25+0,3=0,55\left(mol\right)\\ V_{hh\left(dktc\right)}=n.22,4=0,55.22,4=12,32\left(l\right)\)
\(n_{SO_2}=\dfrac{m}{M}=\dfrac{3,2}{64}=0,05\left(mol\right)\\ n_{N_2}=\dfrac{9,8}{28}=0,35\left(mol\right)\\ n_{hh}=n_{SO_2}+n_{N_2}=0,05+0,35=0,4\left(mol\right)\\ V_{hh}=n.22,4=0,4.22,4=8,96\left(l\right)\)