Б.27.3%
CO2.
ω=Ar(C)/Mr(CO2)·100%=12/44·100%=27.3%
W = (n(ел) * M(ел)) / M(реч)
M(H2O2) = 2 + 32 = 34 г/м
w (O) = 2 * 16 / 34 = 0.94 ; 94%
w(H) = 100 - 94 = 6%
<em>1)Fe+4HNO3=Fe(NO3)3+NO+2H2O</em>
<em>2)Fe+H2SO4=FeSO4+H2</em>
KCLO---K+CL2+O2
O2+Ca=CAO
CAO+H2O=CAOH2+O2
CAOH2+NANO3----CANO3)2+NAOH
а)Fe + HCI-><span>2)FeCI2 + H2</span>
б)Fe(OH)2 + CO2->Fe <span>(HCO3) 2 или 5) <span>FeCO3 + H2O</span></span>
в)Fe(OH)2 + HCI->3)<span> FeCI2 + H2O</span>
<span><span>№2</span></span>
<span><span>а)<span>1. </span>
<span>BeO + Mg = Be + MgO </span>
<span>MgO + 2HCl(разб.) = MgCl2 + H2O </span>
<span>MgCl2 + 2NaOH(разб.) = Mg(OH)2(осадок) + 2NaCl</span></span></span>
Mg(OH)2=MgO+H2O
<span><span>б)С+O2=CO2</span></span>
<span><span>CO2+Na2O=Na2CO3</span></span>
<span><span>Na2Co3+ H2SO4=<span>Na2SO4+H2O+CO2</span></span></span>
<span><span><span><span>Na2SO4+BaCl2=<span>BaSO4(осадок) +2NaCl</span></span></span></span></span>
незабудьте отметить лучшее решение если помогла
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