Question

помогите пожалуйста химия 7 класс
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Answer 1

●Mᵣ(C₂H₆) = 2Aᵣ(C) + 6Aᵣ(H) = 2*12 + 6*1 = 24 + 6 = 30.
$\boxed{\sf \omega (C) = \frac{2A_r(C)}{M_r(C_2H_6)} = \frac{2*12}{30} = \frac{24}{30} = 0,8=80\%}$
Ответ : 80%.
●Mᵣ(K₂CO₃) = 2Aᵣ(K) + Aᵣ(C) + 3Aᵣ(O) = 2*39 + 12 + 3*16 = 78 + 12 + 48 = 138.
$\boxed{\sf \omega (C) = \frac{A_r(C)}{M_r(K_2CO_3)} = \frac{12}{138} \approx 0,087 \approx 8,7\%}$
Ответ : 8,7%.
●Mᵣ((NH₄)₂CO₃) = 2Aᵣ(N) + 8Aᵣ(H) + Aᵣ(C) + 3Aᵣ(O) = 2*14 + 8*1 + 12 + 3*16 = 28 + 8 + 12 + 48 = 96.
$\boxed{\sf \omega (C) = \frac{A_r(C)}{M_r((NH_4)_2 CO_3)} = \frac{12}{96} = 0,125 = 12,5\%}$
Ответ : 12,5%.
●Mᵣ(Al₄C₃) = 4Aᵣ(Al) + 3Aᵣ(C) = 4*27 + 3*12 = 108 + 36 = 144.
$\boxed{\sf \omega(C) = \frac{3A_r(C)}{M_r(Al_4C_3)} = \frac{3*12}{144}= \frac{36}{144}=0,25=25\%}$
Ответ : 25%.