Решение: Скорость реакции H2 + I2 -› 2HI определяем по формуле
н=k*[H2]*[I2]
Следовательно, скорость реакции в начальный момент: н=0,16*0,04*0,05=0,00032=3,2*10-4
v"=0,16*0,03*0,04(так как уменьшилась конц. водорода=>настолько же уменьшится конц. иода)=0,000192 =1,92 · 10 -4
<span>Cu2S + HNO3 = Cu(NO3)2 + SO2 + NO +H2O.</span>
<span>2</span><span>HNO</span>3<span> => 2</span><span>NO</span><span> + </span><span>H</span>2<span>O</span><span> + 3</span><span><u>O</u></span>
<span /><span>Cu</span>2<span>S</span><span>=> 2</span><span>Cu</span><span>(</span><span>NO</span>3<span>)</span>2<span> + </span><span>H</span>2<span>SO</span>4
<span>Cu</span>2<span>S</span><span> + 4</span><span>HNO</span>3<span> => 2</span><span>Cu</span><span>(</span><span>NO</span>3<span>)</span>2<span> + </span><span>H</span>2<span>SO</span>4
<span>5</span><span><u>O</u></span><span> + Cu</span><span>2</span><span>S + 4HNO</span><span>3</span><span> => 2Cu(NO</span><span>3</span><span>)</span><span>2</span><span> + H</span><span>2</span><span>SO</span><span>4</span><span> + H</span><span>2</span><span>O</span>
<span>3<span>Cu</span>2<span>S</span> + 22<span>HNO</span>3 => 6<span>Cu</span>(<span>NO</span>3)2 + 3<span>H</span>2<span>SO</span>4 + 10<span>NO</span> + 8<span>H</span>2<span>O</span></span>
<span><span><span>1) сульфит натрия Na2SO3 - восстановитель.</span></span></span>
<span><span><span>2) сульфид меди Cu2S - восстановитель.</span></span></span>
<span><span><span>3) <span>Нитрат меди(II) Cu(NO3)2</span> - окислитель.</span></span></span>
<span><span><span>4) Оксид азота(<span>II)</span></span></span></span><span> NO - окислитель.</span>
<span>5) Вода H2o - восстановитель.</span>
Mr[Mg₃(PO₄)₂]=24x3+(31+16x4)x2=72 +62+128=262
ω(Mg)=72÷262=0,275 ω%(Mg)=0,275×100%=27,5%
ω(P)=62÷262=0,236 ω%(P)=0,236×100%=23,6%
ω(O)=128÷262=0,489 ω%(O)=0,488×100%=48,9%
1. а) ;
2. в) ;
3. б) ;
4. б) ;
5. 2, С6Н12, С3Н6 т.е. гексен, пропен .
M(раствора)=30,4+349,6=380г
w(FeSO4)=30.4*100%/380=8%
8%+25%=33%
30.4*100/(380-х)г=33
3040=33*(380-х)
3040=12540-33х
33х=12540-3040
33х=9500
х=9500/33
х=287,87878г=288г
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m1(раствора)=380-288=92г
m(Fe2SO4)=30.4г
w(FeSO4)=30.4*100%/92=33%
33%-8%=25%
