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a) V\(_{H2}=0,6.22,4=13,44\left(l\right)\)
b) n\(_{SO3}=\frac{40}{80}=0,5\left(mol\right)\)
V\(_{SO3}=0,5.22,4=11,2\left(l\right)\)
c) n\(_{O2}=\frac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)
V\(_{O2}=0,5.22,4=11,2\left(l\right)\)
d) n\(_{SO2}=\frac{8,8}{44}=0,2\left(mol\right)\)
V\(_{SO2}=0,2.22,4=4,48\left(l\right)\)
e) n\(_{NO2}=\frac{12.10^{23}}{6.10^{23}}=2\left(môl\right)\)
V\(_{NO2}=2.22,4=44,8\left(l\right)\)
f) V\(_{Cl2}=O,75.22,4=16.8\left(l\right)\)
g) n\(_{SO2}=\frac{32}{64}=0,5\left(mol\right)\)
V\(_{SO2}=0,5.22,4=11,2\left(l\right)\)
a) \(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)
b) \(n_{N_2}=\dfrac{1,8.10^{23}}{6.10^{23}}=0,3\left(mol\right)\)
=> \(m_{N_2}=0,3.28=8,4\left(g\right)\)
c) \(n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)=>V_{CO_2}=0,2.22,4=4,48\left(l\right)\)
d) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
=> Số phân tử H2 = 0,15.6.1023 = 0,9.1023
e) \(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
f) \(n_{Cl_2}=\dfrac{3,6.10^{23}}{6.10^{23}}=0,6\left(mol\right)\)
=> VCl2 = 0,6.22,4 = 13,44(l)
g) \(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
=> mO2 = 0,3.32 = 9,6(g)
h) \(n_{K_2O}=\dfrac{18,8}{94}=0,2\left(mol\right)\)
=> Số phân tử K2O = 0,2.6.1023 = 1,2.1023
i) \(n_{CaO}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
=> Số phân tử CaO = 0,2.6.1023 = 1,2.1023
nHCl = 0,2.1,5 = 0,3 (mol)
=> mHCl = 0,3.36,5 = 10,95(g)
\(a_1,m_{CaCO_3}=0,25.100=25(g)\\ a_2,m_{SO_2}=\dfrac{3,36}{22,4}.64=9,6(g)\\ a_3,m_{H_2SO_4}=\dfrac{9.10^{23}}{6.10^{23}}.98=147(g)\)
\(a.V_{N_2}=n.22,4=0,25.22,4=5,6\left(l\right)\)
\(b.n_{NH_3}=\dfrac{0,9.10^{23}}{6.10^{23}}=0,15\left(mol\right)\\ V_{NH_3}=n.22,4=0,15.22,4=3,36\left(l\right)\)
\(c.n_{SO_2}=\dfrac{3,2}{64}=0,05\left(mol\right)\\ V_{SO_2}=n.22,4=0,05.22,4=1,12\left(l\right)\)
\(a,m_{CaSO_4}=136.0,25=34\left(g\right)\\ b,n_{Cu_2O}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\\ m_{Cu_2O}=0,5.144=72\left(g\right)\\ c,n_{NH_3}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ m_{NH_3}=17.0,3=5,1\left(g\right)\\ d,m_{C_4H_{10}}=0,17.58=9,86\left(g\right)\\ e,n_{Cu\left(OH\right)_2}=\dfrac{4,5.10^{25}}{6.10^{23}}=75\left(mol\right)\\ m_{Cu\left(OH\right)_2}=98.75=7350\left(g\right)\\ g,m_{MgO}=0,48.40=19,2\left(g\right)\\ h,n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ m_{CO_2}=44.0,15=6,6\left(g\right)\\ i,m_{Al\left(OH\right)_3}=78.0,25=19,5\left(g\right)\\\)
Các câu còn lại em làm tương tự nha!
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
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) m\(_{K2SO4}=0,3.114=34,2\left(g\right)\)
b) m\(_{AlCl3}=0,4..133,4=53,4\left(g\right)\)
c)n\(_{SO2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
m\(_{SO2}=0,3.64=19,2\left(g\right)\)
d n\(_{O2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)
m\(_{O2}=0,5.32=16\left(g\right)\)
e) m\(_{Ba\left(OH\right)2}=0,75.171=128,25\left(g\right)\)
g) n\(_{ZnSO4}=\frac{10,5.10^{23}}{6.10^{23}0}=1,75\left(mol\right)\)
m\(_{ZnSO4}=1,75.161=281,75\left(g\right)\)
h)n\(_{N2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
m\(_{N2}=0,2.28=5,6\left(g\right)\)
k) m\(_{CaCl2}=0,8.111=88,8\left(g\right)\)
l) m\(_{MgCO3}=0,6.94=50,4\left(g\right)\)