Question

Câu 4: 1,5 . 1023 phân tử chlorine (Cl2) có khối lượng là bao nhiêu? Chiếm thể tích là bao nhiêu lít ở đkc?

(Cl=35,5)

Answer 1

\(n_{Cl_2}=\dfrac{1,5.10^{23}}{6.10^{23}}=0,25\left(mol\right)\\ m_{Cl_2}=n.M=0,25.71=17,75\left(g\right)\\ V_{Cl_2\left(đkc\right)}=n.22,4=0,25.22,4=5,6\left(l\right)\)

Answer 2

\(n_{Cl_2}=\dfrac{1,5.10^{23}}{6.10^{23}}=0,25\left(mol\right)\)

\(m_{Cl_2}=n_{Cl_2}.M_{Cl_2}=0,25.71=17,75\left(g\right)\)

\(V_{Cl_2}=n.24,79=0,25.24,79=6,1975\left(lít\right)\)

Answer 3

nCl2=1,5.10236.1023=0,25(mol)mCl2=n.M=0,25.71=17,75(g)VCl2(đkc)=n.22,4=0,25.22,4=5,6(l)

Answer 4

\(n_{Cl_2}=\dfrac{1,5.10^{23}}{6.10^{23}}=0,25mol\)

\(m_{Cl_2}=0,25.71=17,75g\)

\(V_{Cl_2}\left(đkc\right)=0,25.24,79=6,1975l\)