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Ratkaisuja

Ratkaisuja. 1. Hiilen osuus. C 2 H 5 OH. M=2*12+6+16=46. 2*12 /46 *100% = 52%. CO 2. M=44. 12/ 44 *100 % = 27 %. M = 62. 12/ 62 *100 % = 19 %. H 2 CO 3. Eniten hiiltä on etanolissa (a). 2. Butaanin palaminen. M = 58 C 4 H 10 + 6 ½ O 2 => 4 CO 2 + 5 H 2 O 500g

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Ratkaisuja

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  1. Ratkaisuja

  2. 1. Hiilen osuus C2H5OH M=2*12+6+16=46 2*12 /46 *100% = 52% CO2 M=44 12/ 44 *100 % = 27 % M = 62 12/ 62 *100 % = 19 % H2CO3 Eniten hiiltä on etanolissa (a)

  3. 2. Butaanin palaminen M = 58 C4H10 + 6 ½ O2 => 4 CO2 + 5 H2O 500g 500/58=8.6 mol m 6.5*8.6= 55.9 mol 34.4 mol 43 mol n V 55.9*22.4 l= 1252 litraa savukaasuja 77.4 mol eli 1.7 m3 Ilman tarve = 100 / 21 *1252 litraa = n. 6,0 m3

  4. 3. Glykolin tiheys x 0% 100% 1,12 1,0 1,07 verranto x / 100 = 0,07 / 0,12 => x = 58.3 %

  5. 4. Nimeämiset • N2O3 dityppitrioksidi • Na2HPO4 natriummonovetyfosfaatti • KHCO3 kaliumvetykarbonaatti • MgS Magnesiumsulfidi

  6. 5. Kaavat • kaliumdikromaatti K2Cr2O7 • magnesiumvetysulfaatti Mg(HSO4)2 • ammoniumjodidi NH4I • kalsiumkloridi CaCl2

  7. 6. pH-laskut • 0,2 M rikkihappo pH = -log(0,2) = • 0.4 M etikkahappo pH = -log(0.4*1.8*10-5) = 2.6 • 0.05 M ammoniakki pOH = -log(0.05*1.8*10-5) =3.0 pH = 14- pOH = 11

  8. 7. Atomien rakenne 20 protonia 41-20 =21 neutronia 20 elektronia 41Ca 35Cl- 14C alkuaine nro 20 17 protonia 35-17 =18 neutronia 17+1 = 18 elektronia alkuaine nro 17 6 protonia 14-6 =8 neutronia 6 elektronia alkuaine nro 6

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