Решение на картинке,,,,,,,,,,,
<span><span>2C</span>6<span>H</span>14<span> + 19O</span>2<span> → 12CO</span>2<span> + 14H</span>2<span>O</span></span>
<span>m(C</span>6<span>H</span>14<span>) = 1 кг = 1000 г</span>
<span>Найдем кол-во гексана:</span>
<span>M(С6H14)=12*6+1*14=86 г/моль</span>
<span>n(C</span>6<span>H</span>14<span>) = m(C</span>6<span>H</span>14<span>) / M(C</span>6<span>H</span>14<span>) = 1000 / 86 = 11,63 моль</span>
<span>Найдем кол-во кислорода:</span>
<span><span>n(O</span>2<span>) = (19/2) * n(C</span>6<span>H</span>14<span>) = (19/2) * 11.63 = 110.5 моль</span></span>
<span>Найдем объем кислорода:</span>
<span>V(O</span>2<span>) = n(O</span>2<span>) * V</span>м<span> = 110.5 * 22.4 = 2474 л</span> <span>ω(О</span>2<span>) = 21% = 0.21</span>
<span>Найдем объем воздуха:</span>
<span>V(воздуха) = V(O</span>2<span>) / ω(О</span>2<span>) = 2474 / 0.21 = 11782,9л</span>
<span>1s</span>²<span> 2s</span>²<span> 2p</span>⁴
это элемент 2 периода, IV группы
кислород (O)
Fe+H20>H2fe02 pravilnoye resenie
1) CaCO3+2HNO3--->Ca(NO3)2+H2CO3
CaCO3 ↓+ 2H⁺¹+NO3⁻¹--->Ca⁺²+2NO3⁺¹+2H⁺¹+CO3⁻²
CaCO3--->Ca⁺²+CO3⁻²