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Chữ hợi hẹp nên khó nhìn, bn cho to ra nhé

Cũng ko khác cho lắm

PTHH: Na₂SO₃ + 2HCl ---> 2NaCl + H₂O + SO₂

Ta có: \(n_{SO_2}=\dfrac{6,2}{22,4}\approx0,3\left(mol\right)\)

Theo PT: \(n_{HCl}=2.n_{SO_2}=2.0,3=0,6\left(mol\right)\)

Ta có: \(C_{M_{HCl}}=\dfrac{0,6}{V_{dd_{HCl}}}=0,2M\)

=> \(V_{dd_{HCl}}=3\left(lít\right)\)

câu 1:

a)3Fe+2O₂-->Fe₃O₄            HH

b)4P+5O₂-->2P₂O₅        HH

c)2Al+6HCl-->2AlCl₃+3H₂        Thế

d)2K+2H₂O-->2KOH+H₂         Thế

f) Cu+2AgNO₃-->Cu(NO₃)₂ +2Ag            Thế

\(\left\{{}\begin{matrix}R1=U1^2:P1=220^2:44=1100\Omega\\R2=U2^2:P2=220^2:66=733,33\Omega\end{matrix}\right.\)

Bóng đèn sáng bình thường, vì: \(U=U1=220V\)

\(\left\{{}\begin{matrix}P=UI=220\cdot\left(\dfrac{44}{220}+\dfrac{66}{220}\right)=110W\\Q_{toa}=A=Pt=110\cdot30\cdot60=198000\left(J\right)\end{matrix}\right.\)

\(A1=P1\cdot t1=66\cdot2\cdot30=3960Wh=3,96kWh\)

\(\Rightarrow T=A1\cdot2200=3,96\cdot2200=8712\left(dong\right)\)

\(R_Đ=\dfrac{U_Đ^2}{P_Đ}=\dfrac{220^2}{44}=1100\Omega\)

\(R_q=\dfrac{U_q^2}{P_q}=\dfrac{220^2}{66}=\dfrac{2200}{3}\Omega\)

\(R_{tđ}=\dfrac{R_Đ\cdot R_q}{R_Đ+R_q}=\dfrac{1100\cdot\dfrac{2200}{3}}{1100+\dfrac{2200}{3}}=440\Omega\)

\(\Rightarrow U_Đ=U_q=U_m=440\cdot0,5=220V\)

\(I_{Đđm}=\dfrac{P_Đ}{U_Đ}=\dfrac{44}{220}=0,2A=I_Đ\)

Đèn sáng bình thường.