\(a,n_{CO_2}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{1,2395}{22,4}=\dfrac{2479}{44800}\left(mol\right)\)

\(b,Theo.CTHH:CO_2\\ \Rightarrow n_C=n_{CO_2}=\dfrac{2479}{44800}\left(mol\right)\\ n_O=2n_{CO_2}=2.\dfrac{2479}{44800}=\dfrac{2479}{22400}\left(mol\right)\\ \Rightarrow m_C=n.M=\dfrac{2479}{44800}.12=\dfrac{7437}{12200}\left(g\right)\\ m_O=n.M=\dfrac{2479}{22400}.16=\dfrac{2479}{1400}\left(g\right)\)

\(c,số.phân.tử.CO_2:n.6.10^{23}=\dfrac{2479}{44800}.6.10^{23}=\dfrac{23240625.10^{16}}{7}\)

a) \(n_{CO_2}=\dfrac{V}{22,4}=\dfrac{1,2395}{22,4}=0,05\left(mol\right)\)

b) \(m_{CO_2}=0,05.44=2,2\left(g\right)\)

c) Số phân tử CO2 là: 0,05.6.10²³ = 0,3.10²³ = 3.10²² (phân tử)

a,\(n_{CO_2}=\dfrac{1,2395}{24,79}=0,05\left(mol\right)\)

b, \(m_{CO_2}=n_{CO_2}.M_{CO_2}=0,05.44=2,2\left(g\right)\)

c, Số phân tử CO₂=\(n_{CO_2}.6.10^{23}=0,05.6.10^{23}=3.10^{22}\)

a, \(n_{CO_2}=\dfrac{1,2395}{22,4}=0,055mol\)

\(b,m_{CO_2}=0,055.44=2,42g\)

\(c,Số.phân.tử.CO_2=0,055.6.10^{23}=3,3.10^{22}\) phân tử

a) \(n_{CO_2}=\dfrac{1,2395}{22,4}\approx0,05\left(mol\right)\)

b) \(m_{CO_2}=0,05.44=2,2\left(g\right)\)

a) \(n_{CO_2}=\dfrac{1,2395}{24,79}=0,05\left(mol\right)\)

b) \(m_{CO_2}=0,05.44=2,2\left(g\right)\)

c) Số phân tử CO₂ = 0,05.6.10²³ = 0,3.10²³

á à thi dg k

\(n_{CO_2}=\dfrac{PV}{RT}=\dfrac{0.99\cdot2.479}{0.082\cdot\left(273+25\right)}=0.1\left(mol\right)\)

\(1bar=0.99atm\)

a, \(n_{CO_2}=\dfrac{1,2395}{22,4}\approx0,055mol\)

b, \(m_{CO_2}=0,055.44=2,42g\\ c,Số.phân.tử.CO_2=0,055.6.10^{23}=3,3.10^{22}\left(phân.tử\right)\)

a) \(n_{CO_2}=\dfrac{V}{22,4}=\dfrac{2.479}{22,4}\approx0,11\left(mol\right)\)

b) \(m_{CO_2}=n.M=0,11=4,84\left(g\right)\)

c) Số phân tử CO₂ là : \(0,11.6.10^{23}=0,66.10^{23}\)

a) \(n_{CO_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

b) \(m_{CO_2}=0,1.44=4,4\left(g\right)\)

c) Số phân tử CO₂ = 0,1.6.10²³ = 0,6.10²³

\(M_{CO_2}=12+16.2=44\left(\dfrac{g}{mol}\right)\)

\(M_{CO_2}=12+16.2=44\left(\dfrac{g}{mol}\right)\)

Số mol C O 2 : n =V/22,4= 5,6/22,4 = 0,25 (mol)

Khối lượng của  C O 2 : m = n.M = 0,25.44 = 11 (gam) 

\(n_{CO_2}=\dfrac{6,1975}{24,79}=0,25\left(mol\right)\)

=> \(m_{CO_2}=0,25.44=11\left(g\right)\)

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