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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) \(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)
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)
\(a.V_{N_2}=n.22,4=0,25.22,4=5,6\left(l\right)\)
\(b.n_{NH_3}=\dfrac{0,9.10^{23}}{6.10^{23}}=0,15\left(mol\right)\\ V_{NH_3}=n.22,4=0,15.22,4=3,36\left(l\right)\)
\(c.n_{SO_2}=\dfrac{3,2}{64}=0,05\left(mol\right)\\ V_{SO_2}=n.22,4=0,05.22,4=1,12\left(l\right)\)
a) m\(_{K2SO4}=0,3.114=34,2\left(g\right)\)
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\(_{MgCO3}=0,6.94=50,4\left(g\right)\)
1a) Khối lượng mol của X là :
MX = 22.2 = 44 (g/mol)
mC = \(\frac{44.81,82}{100}\approx36\left(g\right)\)
mH2 = \(\frac{44.18,58}{100}\approx8\left(g\right)\)
nC = 36/12 = 3 (mol)
nH2 = 8/2 = 4 (mol)
Vậy CTHH là C3H4 (Propin).
b) Tương tự câu a.
CTHH là chất khí a là : SO2
2. a) nO2 = 32/16 = 2 mol
b) nO2 = 4,48/22,4 = 0,2 mol
c) nO2 = 3,01.1023/6,02.1023 = 0,5 mol
ta có: nCl2=\(\frac{7,1}{71}=0,1mol\)
\(V_{Cl2}=0,1.22,4=2,24\left(l\right)\)
\(n_{CO2}=\frac{8,8}{44}=0,2\left(mol\right)\)
\(V_{CO2}=0,2.22,4=4,48\left(l\right)\)
\(n_{NO2}=\frac{4,6}{46}=0,1\left(mol\right)\)
\(V_{NO2}=0,1.22,4=2,24\left(l\right)\)
\(n_{h^2}=0,1+0,2+0,1=0,4\left(mol\right)\)
\(V_{h^2}=2,24+2,24+4,48=8,96\left(l\right)\)
b) ta có \(n_{O2}=\frac{16}{32}=0,5\left(mol\right)\)
\(n_{N2}=\frac{14}{28}=0,5\left(mol\right)\)
\(\Leftrightarrow n_{h^2}=0,5+0,5=1\left(mol\right)\)
c) vì \(S=n.6.10^{23}\Rightarrow n=\frac{S}{6.10^{23}}\)
\(n_{N2}=\frac{1,5.10^{23}}{6.10^{23}}=0,25\left(mol\right)\)
\(V_{N2}=0,25.22,4=5,6\left(l\right)\)
\(n_{CO2}=\frac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)
\(V_{CO2}=1,5.22,4=33,6\left(l\right)\)
chúc bạn học tốt like mình nha
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\(_{SO2}=0,5.22,4=11,2\left(l\right)\)