150 * 0,15 = 22,5 г - масса сахара
150 - 22,5 = 127,5 г - масса(объем) воды
22,5 / 8 = 2,8 ложек сахара
42/(14n-6)=57,75/(14n-7+14+32)
n=9
C9H12
1,2,3-триметилбензол или 2, 4, 6-триметилбензол
Первое:
<span><span>2</span><span>Na</span><span>O</span><span>H</span></span><span> + </span><span><span>H</span><span>2</span><span>S</span><span>O</span><span>4</span></span><span> = </span><span>Na</span><span>2</span><span>S</span><span>O</span><span>4</span><span> + </span><span>2</span><span>H</span><span>2</span><span>O</span>
<span><span>2</span><span>Na</span>+<span> + </span><span>2</span><span>O</span><span>H</span>−<span> + </span><span>2</span><span>H</span>+<span> + </span><span>S</span><span>O</span><span>4</span>2−<span> = </span><span>2</span><span>Na</span>+<span> + </span><span>S</span><span>O</span><span>4</span>2−<span> + </span><span>2</span><span>H</span><span>2</span><span>O</span></span>
<span><span><span>2</span><span>O</span><span>H</span>−<span> + </span><span>2</span><span>H</span>+<span> = </span><span>2</span><span>H</span><span>2</span><span>O</span></span></span>
<span><span><span>Второе:</span></span></span>
<span><span><span>HNO3 + K2CO3 = HK2 + CO3NO3</span></span></span>
<span><span><span><span>HNO</span>3<span> + K</span>2<span>CO</span>3<span> = HK</span>2<span> + CO</span>3<span>NO</span>3</span></span></span>
Третье:
2HCl+MgО=MgCL2+H2O
2H(+)+2Cl(-)+MgО=Mg(+2)+2CL(-)+H2O
2H(+)+MgО=Mg(+2)+H2O
Четвёртое:
<span><span>3</span><span>K</span><span>O</span><span>H</span></span><span> + </span><span><span>Fe</span>(<span>N</span><span>O</span><span>3</span>)<span>3</span></span><span> = </span><span>3</span><span>K</span><span>N</span><span>O</span><span>3</span><span> + </span><span>Fe</span><span>(<span>O</span><span>H</span>)</span><span>3</span>
<span><span>3</span><span>K</span>+<span> + </span><span>3</span><span>O</span><span>H</span>−<span> + </span><span>Fe</span>3+<span> + </span><span>3</span><span>N</span><span>O</span><span>3</span>−<span> = </span><span>3</span><span>K</span>+<span> + </span><span>Fe</span><span>(<span>O</span><span>H</span>)</span><span>3</span><span>+ </span><span>3</span><span>N</span><span>O</span><span>3</span>−</span>
<span><span>3</span><span>O</span><span>H</span>−<span> + </span><span>Fe</span>3+<span> = </span><span>Fe</span><span>(<span>O</span><span>H</span>)</span><span>3</span><span>+</span></span>
Пятое хз каак..
Шестое:
ZnSO4 +K3PO4=ZnSO4<span> +KOH</span>
Раз четвертая группа, значит высший оксид типа ХО2
M(XO2) = (2*M(O))/w(O) = (2*16)/0,4 = 80
M(X) = M(XO2) - 2 * M(O) = 80 - (2 * 16) = 48 - титан (Ti)
в реакции коэфциенты везде 1 так что получается что 2 моль водорода.
2*22,4 = 44,8л