Rainer Steiger

Soluzione:

1. · Soluzione di sulfato di rame: CuSO₄, rispettivamente Cu²⁺/SO2-4
2.     Cu²⁺ + 2e^– → Cu, z = 2
3. · m/M = I·t/(z·F)
4.     m = I·t·M/(z·F) = 5.8·3600·3·63.5/(2·96485) = 20.6 g
5.     Unità: [A·s·g/mol/(A·s/mol)] = [g]
6. · Es fabbricazione approximativamente: il volumi (!) il strato di sfera è uguale il superficie di sfera moltiplicato con il altezza, oppure Magre (:d) il scatto V = 4·π·r²·d
7. · Densità = m/V
8.     m = densità · V = Dichte·4·π·r²·d
9.    
10.     → d = m/(Dichte·4·π·r²)
11.     d = 20.63/(8.9·4·π·8.3²) = 2.67·10^-3cm = 0.026 mm

Solution:

Notes:
— A first step is to consider which ions the salt 'melt' into, balance it, and then define the solubility product.
— The polyatomic anions – cations recognition. 'disintegration' not continued.
— The balanced reaction for an (!) educated part formulation!

1. NaCl ⇄ Na⁺ + Cl^–      L_p(NaCl) = c(Na⁺)·c(Cl^–)
2. CaCO₃ ⇄ Ca²⁺ + CO2-3      L_p(CaCO₃) = c(Ca²⁺)·c(CO2-3)
3. CaF₂ ⇄ Ca²⁺ + 2F^–      L_p(CaF₂) = c(Ca²⁺)·c²(F^–)
4. NH₄Cl ⇄ NH+4 + Cl^–      L_p(NH₄Cl) = c(NH+4)·c(Cl^–)
5. AlBr₃ ⇄ Al³⁺ + 3 Br^–      L_p(AlBr₃) = c(Al³⁺)·c³(Br^–)
6. Fe₂O₃ ⇄ 2Fe³⁺ + 3O^2-      L_p(Fe₂O₃) = c²(Fe³⁺)·c³(O^2-)
7. CaSO₄ ⇄ Ca²⁺ + SO2-4      L_p(CaSO₄) = c(Ca²⁺)·c(SO2-4)

Solution:

1. 2·H₃O⁺ + 2·e^– → H₂ + 2·H₂O // – 2·H⁺
2. 2·H₂O + 2·e^– → H₂ + 2·OH^– // ·2
3. 4·H₂O + 4·e^– → 2·H₂ + 4·OH^–

Soluzione:

1. · Il volume della superficie da arricchire è:
2.     V= O·h = 120cm²·0.02cm = 2.4cm³
3. · Con densità = m/V:
4.     m=densità·V= 10.5/cm³ · 2.4 cm³ = 25.2 grammi
5. · Nitrato di argento: AgNO₃, cioè Ag⁺/NO–3
6.     Ag⁺ + 1e^– → Ag, z= 1
7. · m/M = I·t/(z·F)
8.     t = m·z·F/(M·I) = 25.2·1·96485/(107.9·10)
9.     t = 2253 s , circa 40 minuti

Solution:

1. 1 M CuSO₄ means that 1 mol copper sulphate is dissolved per litre (solvent, very commonly water).
2. Manufacture: 1 mol weighs 159.6 grams. So 159.6 grams of copper sulfate are weighed and added to the water until the 1 litre mark is reached. The own volume of copper sulphate need not be taken into account.
3. Approximately 1 liter of water is needed = 1000 grams. This would be (1000/18) 55.5 mol water. So the ratio of ions is:
       Cu²⁺ to SO24 to water = 1 : 1 : 55.5

Solution:

Consider the reverse case: dissolving ions from water. Since there is an ion-dipole interaction between these two particles, a force must be used to separate them. Now we actually have a reverse process, namely, the attachment of water to an ion. From a logical point of view, for the reverse case ('solving a dipole ion interaction') the same 'volume' of energy must be released as when attaching an ion to a dipole. The only difference is the energy signal. The energy itself is called hydration energy.

Solution:

1. 4·OH^– → O₂ + 2·H₂O ou plus précisément:
2. 4·OH^– → O₂ + 2·H₂O + 4·e^– // [traduction manquante] 4H⁺
3. 4·OH^– + 4H⁺ → O₂ + 2·H₂O + 2H⁺ 2H⁺ + 4·e^–
4. 4·H₂O → O₂ + 2·H₃O⁺ + 2H⁺ + 4·e^– // [traduction manquante] 2·H₂O
5. 6·H₂O → O₂ + 2·H₃O⁺ + (2H⁺ + 2·H₂O) + 4·e^–
6. 6·H₂O → O₂ + 2·H₃O⁺ + ( 2·H₃O⁺ ) + 4·e^–
7. 6·H₂O → O₂ + 4·H₃O⁺ + 4·e^–

Soluzione:

1. · 10 A = 10 C/s
2. · Numero di elettroni al secondo:
3.     10/1.602·10^-19 = 6.24·10¹⁹
4. · Complesso Masse(e^–) per seconda:
5.     6.24·10¹⁹·9.1E.31 = 5.68·10^-11 kg
6. · Da quanto a 1 kg totale:
7.     1/5.68·10^-11 = 1.76·10¹⁰ s = circa 558 anni

Solution:

In 1 mol/L NaOH se passe avec à base de métal l'argent (Ag) à partir de de façon normale conditions générales aucun réaction d’oxydoréduction.
L'argent est trop précieux: OH^– (et l'eau) ne sont pas oxydant qui pourrait oxider Ag à Ag⁺.

Solution:

Water can itself to a ion to be invested, these layers to the ion is hydration shell called. Note: there is a difference in the spatial arrangement of the water molecule when a cation or anion is present.

Soluzione:

1. CuCl₂: ioni Cu²⁺ oppure Cl^–
       Cu²⁺ + 2e^– → Cu, z = 2
2. m/M = I·t/(z·F)
       t = m·z·F/(M·I) = 3.5·2·96485/(63.546·0.45)
       t = 23618 s = circa 6.6 h

Soluzione:

Nota: 35 µm corrispondono a 35·10^-6m

1. Il volume dello strato di cromo da applicare è:
       V= O·h = 15m^2·35·10-6m = 0.000525 m³ = 525 cm³
2. Con densità = m/V:
       m=densità·V= 7.15g/cm³ · 525 cm³ = 3753.75 grammi
3. CrO₃, coinvolti ioni: Cr⁶⁺/O^2-
       Cr⁶⁺ + 6e^– → Cr, z = 6
4. m/M = I·t/(z·F)
       t = m·z·F/(M·I) = 96485·6·3753/(52·450)
       t = 92848 s (a 100% di efficacia)
5. 90% di efficacia: 103164 secondi

Solution:

1. F_c = k·Q₁·Q₂/r²
       Q₁ = Q₂ = elementary load = 1.60222·10^-19 C
       F_c = (8.9875·10⁹)·(1.6022·10^-19)² / r²
       F_c = 2.31·10^-28 / r² [N]
2. F_G = G·m₁·m₂/r²
       · Individual atoms take … u (unit)
       · 1 u = 1.66·10^-27 kg
       · m(Na⁺) = 22.99 u
       · m(Cl^–) = 35.45 u
       · F_G = 6.67·10^-11·22.99u·35.45u / r²
       · F_G = 1.5·10^-61 / r² [N]
3. Factor x = F_c/F_G
       x = (2.31·10^-28 / r² ) / (1.5·10^-61 / r² ) = 1.5·10³³

The distance between the ions shrinks in the equation, and as a result, it can be said that the Coulomb force is clearly stronger than the gravitational force.

Solution:

1. Sn + 2H₃O⁺ → Sn²⁺ + H₂ + 2·H₂O
2. Sn → Sn²⁺ , oxygène 0 → +Ⅱ, oxydation
3. Au lieu de H₃O⁺: H₂O + H⁺
       2·H⁺ → H₂ , nombre d'ox +Ⅰ → 0, réduction
4. Oui, les nombres d’oxydation le acide prendre de +Ⅰ sur O à partir de:
       Un oxydant a un effet oxydant

Soluzione:

Nitrato d'argento: AgNO₃, metallione: Ag⁺
Nota 1: m/M = I·t/(z·F)
Nota 2: In un primo passo, assumere un grado di efficacia del 100%.
Da m/M = I·t/(z·F) segue:
m = (I·t·M)/(z·F) = (2.54·45·60·107.87) / (1·96485) = 7.67 g
Con efficacia del 90%: 6.9 grammi
Unità: [(A·s·g/mol)/(C/mol)], [A] = [C/s]

Solution:

c=n/V=m/M/v      –> m =c·V·M = 0.2 mol/l·0.4 l · 58.5 g/mol = 4.68 g
Manufacture: 4.68 g of NaCl is put into a glass and then filled with water up to 0.4 litres.
The answer: it's going to be added 0.4 liters now is not entirely accurate, because this is not (at least) the same amount as filling up to the 0.4 liters mark: NaCl takes a small volume.

Solution:

1. C'est vrai
2. C'est vrai
3. Faux, plus à droite, plus l'oxydant est puissant.
4. C'est faux, c'est vrai: les réducteurs émettent des électrons
5. Faux. correct serait: la rouille est une forme particulière de corrosion du fer ou
       Tous les métaux peuvent être oxydés, mais tous ne peuvent pas se rouiller.
6. C'est correct: une substance est oxydée quand elle émet des électrons.
7. C'est vrai. Par définition, l'anode est le lieu de l'oxydation (décharge électronique), ce qui est toujours le cas, qu'il s'agisse d'une cellule galvanique ou d'une électrolyse.
8. C'est faux. C'est correct: … a une forte tendance à absorber les électrons
9. C'est vrai.

Soluzione:

Da m/M = I·t/(z·F) segue:
m = (I·t·M)/(z·F) = (2.5·30·60·63.546) / (2·96485) = 1.48 g
Con livello di efficacia 92.5%: 1.37 grammi
Unità: [(A·s·g/mol)/(C/mol)], [A] = [C/s]

Solution:

Cl(0)2 + 2Na(+i)O(-ii)H(i) -> Na(+i)Cl(-i) + Na(+i)Cl(i)O(-ii) + H(+i)2O(-ii)

{Cl: } {0} -> {-I} {: Réduction}

{Cl: } {0} -> {+I} {: oxydation}

{H: } {+I} -> {+I} {: aucune modification}

{Na: } {+I} -> {+I} {: aucune modification}

{O: } {-II} -> {-II} {: aucune modification}

On observe donc que le chlore est oxydé et réduit en même temps. In d’autres un le cas est de une disproportionnement parlant. Et pour les spécialistes, la réaction inverse est appelée coproportionnement ou synoproportionnement.

Solution:

This is the polyatomic anions:

1. Sulphate: SO2-4
2. Nitrate: NO–3
3. Carbonate: CO2-3
4. Hydrogen sulphate: HSO–4
5. Phosphate: PO3-4

The Lewis structures you can see in the video. Nothing done yet.

Solution:

{a: } 3S(+iv)O(-ii)3^2- + Cr(+vi)2O(-ii)7^2- + 8H(+i)^+ -> 2Cr(+iii)^3+ + 3S(+vi)O(-ii)4^2- + 4H(+i)2O(-ii)

$itemColor1(white){à vide} {S: } {IV} -> {VI} {: oxydation}

$itemColor1(white){à vide} {Cr: } {VI} -> {III} {: Réduction}

$itemColor1(white){à vide} {H: } {+I} -> {+I} {: aucune modification}

$itemColor1(white){à vide} {O: } {-II} -> {-II} {: aucune modification}

{b:} N(-ii)2H(+i)4 + 4Fe(+iii)(C(+ii)N(-iii))6^3- + 4O(-ii)H(+i)^- -> N(0)2 + 4Fe(+ii)(C(+ii)N(-iii))6^4- + 4H(+i)2O(-ii)

$itemColor1(white){à vide} {N: } {-II} -> {0} {: oxydation}

$itemColor1(white){à vide} {Fe: } {+III} -> {+II} {: Réduction}

$itemColor1(white){à vide} {H: } {+I} -> {+I} {: aucune modification}

$itemColor1(white){à vide} {C: } {+II} -> {+II} {: aucune modification}

$itemColor1(white){à vide} {O: } {-II} -> {-II} {: aucune modification}

{c:} Ru(0) + 6Cl(-i)^- + 3N(+v)O(-ii)3^- + 6H(+i)^+ -> Ru(+iii)Cl(-i)6^3- + 3N(+iv)O(-ii)2 + 3H(+i)2O(-ii)

$itemColor1(white){à vide} {Ru: } {0} -> {+III} {: oxydation}

$itemColor1(white){à vide} {N: } {+V} -> {+IV} {: Réduction}

$itemColor1(white){à vide} {Cl: } {-I} -> {-I} {: aucune modification}

$itemColor1(white){à vide} {H: } {+I} -> {+I} {: aucune modification}

$itemColor1(white){à vide} {O: } {-II} -> {-II} {: aucune modification}

{d:} Cl(+v)O3(-ii)^- + 3S(+iv)O2(-ii) + 3H2(+i)O(-ii) -> 3S(+vi)O4(-ii)^2- + Cl(-i)^- + 6H(+i)^+

$itemColor1(white){à vide} {Cl: } {+V} -> {-I} {: Réduction}

$itemColor1(white){à vide} {S: } {+IV} -> {+VI} {: oxydation}

$itemColor1(white){à vide} {H: } {+I} -> {+I} {: aucune modification}

$itemColor1(white){à vide} {O: } {-II} -> {-II} {: aucune modification}

{e:} 3H(+i)2S(-ii) + 2H(+i)N(+v)O(-ii)3 -> 2N(+ii)O(-ii) + 3S(0) + 4H(+i)2O(-ii)

$itemColor1(white){à vide} {S: } {-II} -> {0} {: oxydation}

$itemColor1(white){à vide} {N: } {+V} -> {+III} {: Réduction}

$itemColor1(white){à vide} {H: } {+I} -> {+I} {: aucune modification}

$itemColor1(white){à vide} {O: } {-II} -> {-II} {: aucune modification}

{f:} K(+i)S(-ii)C(iv)N(-iii) + 4H(+i)2O(-iii) + 4I(0)2 -> K(+i)H(+i)S(+vi)O(-ii)4 + 7H(+i)I(-i) + I(+i)C(+ii)N(-iii)

$itemColor1(white){à vide} {S: } {-II} -> {+VI} {: oxydation}

$itemColor1(white){à vide} {C: } {+IV} -> {+II} {: Réduction}

$itemColor1(white){à vide} {I: } {O} -> {-I} {: Réduction}

$itemColor1(white){à vide} {K: } {+I} -> {+I} {: aucune modification}

$itemColor1(white){à vide} {H: } {+I} -> {+I} {: aucune modification}

$itemColor1(white){à vide} {O: } {-II} -> {-II} {: aucune modification}

Solution:

There is no known compound with the chemical formula MgO₄. Magnesium usually forms compounds with an oxidation number of +2, which means that magnesium ions (Mg²⁺⁾ are present. Therefore, the compound magnesium oxide (MgO) should be expected.

Additionally, in MgO₄ the magnesium would have to have a charge of 8 +, since the oxygen ions do have a charge of -2 each. A look at the usual charges shows that somewhere at a charge of about 5 + the atom is so positively charged that it absorbs electrons no matter where and thus reduces its (positive) charge.

Quintessence: A purchase is not worth it.

Solution:

There are thousands of salts. chatgpt's answer, including its eloquence:
Salts are compounds consisting of positively charged metallions and negatively charged nonmetallions (anions). Here are ten examples of typical salts:

1. Sodium chloride (NaCl): Ordinary salt used in the kitchen.
2. Potassium nitrate (KNO₃): A salt used in the food industry and as a fertilizer.
3. Calcium carbonate (CaCO₃): This salt is found in limestone, chalk and marble.
4. Eisen(II)-sulphate (FeSO₄): a salt, the often as fertilizers and to the water treatment used is.
5. Sodium carbonate (Na₂CO₃): Also known as soda, is used in glassmaking and as a cleaning agent.
6. Sodium sulphate (Na₂SO₄): A salt used as a drying agent and in soap making.
7. Aluminium oxide (Al₂O₃): This salt occurs naturally as bauxite and is used for aluminium extraction.
8. Ammonium chloride (NH₄Cl): A salt used in some acid-base reactions and in medicine.
9. Sodium hydrocarbonate (NaHCO₃): Also known as sodium, is used in baking powder.
10. Calcium chloride (CaCl₂): A salt used as a food additive and as a wintering agent.

These examples illustrate the variety of salts used in various fields such as food industry, agriculture, chemistry and medicine.
Commentary: the names including the salt formulas are flawless. Quintessence: The typical (current) invention of facts by chatgpt could not be observed. A check with other sources is good and important.

Soluzione:

Una miscela di (almeno) due metalli è chiamata lega. Idealmente, la miscela ha una proprietà migliore e più pratica del metallo puro stesso. Z.B. È una miscela su fabbricazione di rame e zinc (fabbricazione di plastica) più dura come le purissimi elementi il proprio.

Solution:

{} 3S(+iv)O3(-ii)^2- + Cr(+vi)2O7(-ii)^2- + 8H(+i)^+ -> 2Cr(+iii)^3+ + 3S(+vi)O4(-ii)^2- + 4H(+i)2O(-ii)

{S: } {+IV} -> {+VI} {L’oxydation}

{Cr: } {+VI} -> {+III} {Réduction}

{H: } {+I} -> {+I} {Aucune modification}

Soluzione:

Solo gli elementi ferro, nichel e cobalto sono magnetici. Anche alcune metalli rari della terra come neodymi, fabbricazione di acciaio e disprosio possono magnetico essere, spesso ma solo a determinati temperature o in allievi, come z.B. In i forti magnete di neodymi.
Quindi, di circa 100 elementi appartenenti ai metalli, circa il 5% sono magnetici. Il magnetismo non è una caratteristica tipica, ma piuttosto un'eccezione.

Solution:

Ag + 2HNO3 -> AgNO3 + NO2 + H2O

Ag(0) + 2H(+i)N(+v)O3(-ii) -> Ag(+i)N(+v)O(-ii)3 + N(+iv)O(-ii)2 + H(+i)2O(-ii)

{Ag: } {0} -> {+I} {L’oxydation}

{N: } {+V} -> {+IV} {Réduction}

{H: } {+I} -> {+I} {Aucune modification}

{O: } {-II} -> {-II} {Aucune modification}

Solution:

1. NaCl: sodium chloride
2. F^–: Fluoride
3. SO2-4 : sulphate
4. O^2-: oxide
5. NaNO₃: Sodium nitrate
6. Al₂O₃: Aluminium oxide
7. FeO: iron (II) oxide
8. Fe₂O₃: iron and steel(III)oxide
9. NaS: Sodium sulphide
10. C^4-: Carbide
11. CaO₂: Calcium oxide
12. NaHCO₃^–: sodium hydrocarbonate
13. Na₂SO₄: Sodium sulfate
14. NH₄⁺: [translation unavailable]
15. PbS: Blei(II)-sulfide
16. Al₄C₃: Aluminium carbide
17. (NH₄)(NO₃): ammonium nitrate
18. AlBr₃ : Aluminium bromide
19. Fe(NO₃)₃ : iron-III nitrate
20. (NH₄)₂S: Ammonium sulphide
21. Fe₃(PO₄)₂ : iron-II phosphate
22. Fe(HPO₄): iron-II hydrogen phosphate
23. Fe(H₂PO₄)₃ : iron-III-dihydrogen phosphate
24. Fe₂(SO₄)₃ : iron-III sulphate
25. Fe(HSO4)₂ : iron-II hydrogen sulphate

Solution:

1. NaCl: Na⁺ / Cl^–
2. Al₂O₃: Al³⁺ / O^2-
3. MgO: Mg²⁺ / O^2-
4. KBr: K⁺ / Br^–
5. FeO: Fe²⁺ / O^2-
6. Fe₂O₃: Fe³⁺ / O^2-
7. Na₂O: Na⁺ / O^2-
8. CaO: Ca²⁺ / O^2-
9. (NH₄)₂HPO₄ : NH+4 / HPO2-4
10. Ca(H₂PO₄)₂ : Ca²⁺ / H₂PO–4
11. Fe(NO₃)₃ : Fe³⁺ / NO–3
12. Ba(ClO₃)₂ : Ba²⁺ / ClO–3
13. Ca(OH)₂: Ca²⁺ / OH^–
14. Ca₃(PO₄)₂ : Ca²⁺ / PO3-4
15. Zn(HCO₃)₂ : Zn²⁺ / HCO–3
16. LiCO₃ : Li⁺ / CO2-3
17. Na₂H₂P₂O₇ : Na⁺ / H₂P₂O2+7

Solution:

1. Les métaux sont-ils toujours des réducteurs ?
       Notez que le discours de métaux est
       Donc z.B… Pour moi → Me⁺
       Résistants: réducteurs et oxydants
       Mg → Mg²⁺ + 2e^–
       Al → Al³⁺ + 3e^–
2. Les métaux sont-ils toujours des oxydants ?
       Remarquez que l'on parle de métallions
       Oxydants: oxydants et réduits
       Mg²⁺ + 2e^– → Mg
       Al³⁺ + 3e^– → Al
       Cu²⁺ + 2e^– → Cu
       ABER … Cu⁺ peut (il n'y a pas après [traduction manquante]) à Cu²⁺ à l'oxydation ou à Cu réduit sont.

Soluzione:

1. Conduzione elettrica
2. Capacità di conduzione termico
3. Deformabilità
4. Brillantezza

Solution:

électrolyse: 2·H₂O ⇄ 2·H₂ + 1·O₂
Déterminer tous les nombres d’oxydation, de l' réactif au produit:
H: +Ⅰ –> 0
O: -Ⅱ –> 0
Les nombres d’oxydation changent, c'est-à-dire qu'il s'agit ici d'une réaction d’oxydoréduction.

Solution:

1. Na + ½ Cl₂ –> NaCl
2. Mg + ½ O₂ –> MgO
3. K + ½ Br₂ –> KBr
4. 4·Al + 3·O₂ –> 2·Al₂O₃
5. Fe + ½ O₂ –> FeO
6. 2·Fe + 3/2·O₂ –> Fe₂O₃
7. 2·Na + ½ O₂ –> Na₂O
8. Ca + ½ O₂ –> CaO

Soluzione:

Le definizioni seguono:

1. Esso vale: 1·CaCO₃ ⇄ 1·Ca²⁺ + 1·CO2-3
2. Quindi vale anche: c(CaCO₃) = c(Ca²⁺) = c(CO2-3)
3. Esso seguono: da: L_p(CaCO₃) = c(Ca²⁺) · c(CO2-3)
       E perché z.B. (Ca²⁺⁾ = c(Ca²⁺)
       L_p(CaCO₃) = c(Ca²⁺) · c(Ca²⁺) = c²(Ca²⁺)     o
       L_p(CaCO₃) = c(CO2-3) · c(CO2-3) = c²(CO2-3)     o
       L_p(CaCO₃) = c(CaCO₃) · c(CaCO₃) = c²(CaCO₃)
4. → c(CaCO₃) = √(L_p) = = √ L_p = 6.93·10^-5 mol/l

Ora c=n/V=m/M/V può essere chiuso su 'm':

m = c(CaCO₃)·M(CaCO₃)·V = 6.93·10^-5 mol/l · 100.1 g/mol · 200 l = 1.39 grammi

Considerato l'enorme quantità d'acqua, poco calcio si dissolve, circa 1.4 grammi.

Solution:

Name of the salt	cation	Lewis structure of the manufacture from materials	salt formula
Sodium sulphate	Na+	SO2-4	Na2SO4
Potassium nitrate	K+	NO–3	KNO3
Calcium phosphate	Ca2+	PO3-4	Ca3(PO4)2
Magnesium carbonate	Mg2+	CO2-3	MgCO3
Ammonium chloride	NH+4	Cl–	NH4Cl
Blei(II)-sulfide	Pb2+	S2-	PbS
Eisen(II)-oxide	Fe2+	O2-	FeO
Eisen(III)-oxide	Fe3+	O2-	Fe2O3
Aluminium carbide	Al3+	C4-	Al4C3
Potassium permanganate	K+	MnO–4	KMnO4
Sodium acetate	Na+	CH3COO–	NaCH3COO
Aluminium oxide	Al3+	O2-	Al2O3

Soluzione:

Applicazione:
— 1·NaCl ⇄ 1·Na⁺ + 1·Cl^–
— c(NaCl) = c(Na⁺) = c(Cl^–)
— L_p(NaCl) = 18.26 mol²/l² = c(Na⁺)·c(Cl^–) = c²(Na⁺) = c²(Cl^–) = c²(NaCl)
–> c(NaCl) = √(L_p) = 4.27 mol/l
Da c=n/V= m/M/V      seguono:: m = c·V·M
per 1.0 litri: m(NaCl) = c(NaCl) ·V·M(NaCl) = 4.27 mol/l·1.0 l · 58.5 g/mol = 249.8 g
per 0.3 litri: m(NaCl) = c(NaCl) ·V·M(NaCl) = 4.27 mol/l·0.3 l · 58.5 g/mol = 74.9 g

Solution:

1. CH₄ + 2·O₂ → 2·H₂O + CO₂
2. La transition de l'réactif au produit est répertoriée.
       H: +Ⅰ → +Ⅰ
       C: -Ⅳ → +Ⅳ
       O: 0 → O dans H₂O: -Ⅱ       O dans CO₂: -Ⅱ
3. L'hydrogène: les oxygènes du H ne changent pas
       Le carbone: changement de -Ⅳ à +Ⅳ: oxydation
       Oxygène: changement de 0 à -Ⅱ: réduction
4. Les oxygènes de l'oxygène vont de 0 (en O₂) à -Ⅱ. Le O₂ est donc réduit, est ici donc le oxydant.

Soluzione:

Nota le partecipanti ioni sono Al³⁺ oppure O^2-

1. Al₂O₃ ⇄ 2·Al³⁺ + 3·O^2-
2. L_p = c²(Al³⁺)·c³(O^2-)
3. Per il numero di particelle, nota '=':
       c(Al₂O₃) = ½·c(Al³⁺) = ⅓·c(O^2-)      
       c(Al³⁺) = 2·c(Al₂O₃)       e:
       c(O^2-) = 3·c(Al₂O₃)       quindi vale:
4. L_p = [2·c(Al₂O₃)]² · [3·c(Al₂O₃)]³
5. L_p = 4·27·c²(Al₂O₃)·c³(Al₂O₃) = 108·c⁵(Al₂O₃)
6. c(Al₂O₃) = 5te radice di (L_p/108) = 5te radice di (1·10^-33/108) = 9.8·10^-8 mol/l
7. Da c=n/V=m/M/V segue …
       m = c·V·M = 9.8·10^-8 mol/L · 1.0 L · 102 g/mol = 9.9·10^-6 g = 1·10^-5 g

Es dissolvere si quindi maggiore circa 0.00001 grammi acido di alluminio in un litri acqua. Il sale è davvero molto poco solubile in acqua.

Solution:

	Na+	NH+4	Fe2+	Mn2+	Al3+	Fe3+
Br–	NaBr	NH4Br	FeBr2	MnBr2	AlBr3	FeBr3
NO3–	NaNO3	(NH4)(NO3)	Fe(NO3)2	Mn(NO3)2	Al(NO3)3	Fe(NO3)3
H2PO4–	NaH2PO4	(NH4)(H2PO4)	Fe(H2PO4)2	Mn(H2PO4)2	Al(H2PO4)3	Fe(H2PO4)3
O2-	Na2O	(NH4)2O	FeO	MnO	Al2O3	Fe2O3
S2-	Na2S	(NH4)2S	FeS	MnS	Al2S3	Fe2S3
SO2-4	Na2SO4	(NH4)2(SO4)	FeSO4	MnSO4	Al2(SO4)3	Fe2(SO4)3
CO2-3	Na2CO3	(NH4)2(CO3)	FeCO3	MnCO3	Al2(CO3)3	Fe2(CO3)3
PO3-4	Na3PO4	(NH4)3PO4	Fe3(PO4)2	Mn3(PO4)2	AlPO4	FePO4

Solution:

1. Réduction: les électrons sont absorbés, le nombre d'oxydations diminue, “réduit”
2. Oxydation: les électrons sont émis, le nombre d'oxydations augmente

Solution:

	Al3+	NH+4	Fe2+	Fe3+	Na+	Mn2+
O2-	Al2O3	(NH4)2O	FeO	Fe2O3	Na2O	MnO
Br–	AlBr3	NH4Br	FeBr2	FeBr3	NaBr	MnBr2
S2-	Al2S3	(NH4)2S	FeS	Fe2S3	Na2S	MnS
NO3–	Al(NO3)3	(NH4)(NO3)	Fe(NO3)2	Fe(NO3)3	NaNO3	Mn(NO3)2
SO2-4	Al2(SO4)3	(NH4)2(SO4)	FeSO4	Fe2(SO4)3	Na2SO4	MnSO4
H2PO4–	Al(H2PO4)3	(NH4)(H2PO4)	Fe(H2PO4)2	Fe(H2PO4)3	NaH2PO4	Mn(H2PO4)2
CO2-3	Al2(CO3)3	(NH4)2(CO3)	FeCO3	Fe2(CO3)3	Na2CO3	MnCO3
PO3-4	AlPO4	(NH4)3PO4	Fe3(PO4)2	FePO4	Na3PO4	Mn3(PO4)2

Solution:

1. Réduction: le substance lui-même absorbe les électrons
2. Oxydation: le substance lui-même émet des électrons
3. Réducteur: le substance lui-même pressé un autres (!!) substance électrons sur, fonctionne donc réducteur sur les autres substance. Le substance lui-même perdue donc électrons, donne donc électrons à partir de, est donc à l'oxydation.
   En résumé: fonctionne réducteur et est à l'oxydation.
4. Oxydant: le substance lui-même prend d’un autres (!!) substance électrons sur, fonctionne donc à oxydation sur les autres substance. Le substance lui-même prend donc électrons sur, est donc réduit.
   En résumé: fonctionne à oxydation et est réduit.

Soluzione:

Nota le partecipanti ioni sono Al³⁺ oppure F^–

1. AlF₃ ⇄ Al³⁺ + 3·F^–
2. L_p = c(Al³⁺)·c³(F^–)
3. Per il numero di particelle, nota '=':
       c(AlF₃) = c(Al³⁺) = ⅓·c(F^–)      
       3·c(AlF₃) = 3·c(Al³⁺) = c(F^–)       quindi vale:
       c(Al³⁺) = c(AlF₃)       e:
       c(F^–) = 3·c(AlF₃)
4. L_p = c(Al³⁺)·c³(F^–) = c(AlF₃)·c³(3·c(AlF₃)) = 27·c⁴(c(AlF₃)
5. c(AlF₃) = ∜(L_p/27) = 0.00123 mol/L
6. Da c=n/V=m/M/V segue …
       m = c·V·M = 1.23·10^-3 mol/L · 1 L · 84 g/mol = 0.1038 g

Es dissolvere si quindi maggiore circa 0.1 grammi fluoruro di alluminio in un litri acqua. Il sale è poco solubile in acqua.

Soluzione:

Nota le partecipanti ioni sono Mg²⁺ oppure Br^–

1. 1·MgBr₂ ⇄ 1·Mg²⁺ + 2·Br^–
2. Si applica quindi 2·c(MgBr₂) = 2·c(Mg²⁺) = c(Br^–)
3. a) se: c(Br^–) = 2·c(Mg²⁺)
4.     L_p = c(Mg²⁺)·c²(Br^–) = c(Mg²⁺)·(2c(Mg²⁺))² = 4c³(Mg²⁺)
5.     c(Mg²⁺) = c(MgBr₂) = ∛(L_p/4) = 0.055 mol/L
6.     m = c·V·M = 0.055mol/l · 1.0 L · 184.113 g/mol =10.24 g/L
7.    
8. b) se c(Mg²⁺) = c(MgBr₂) = ½·c(Br^–)
9.     L_p = ½·c(Br^–) ·c²(Br^–) = ½·c³(Br^–)
       oppure c(Br^–) =∛(2·L_p)
10.     c(Br^–) = 2c(MgBr₂)
11. quindi c(MgBr₂) = ½·c(Br^–) = ½·∛(2·L_p)
        = ∛(1/8)·∛(2·L_p) = ∛(2·L_p/8) = ∛(L_p/4)
12.     Si ottiene la stessa soluzione di a: c(MgBr₂) = ∛(L_p/4) = 0.055 mol/L
13.     Quindi il resto sarebbe identico e quindi m=10.24 g/L

Solution:

Start-Tabelle-2

a) H2		H: 0
b) F2		F: 0
c) Na+		Na+: +Ⅰ
d) S2-		S2-: -Ⅱ
e) H2O		H: +Ⅰ	O: -Ⅱ
f) Fe3O4		Fe: +Ⅲ	O: -Ⅱ
g) CH4		C: -Ⅳ	H: +Ⅰ
h) CH3F		F: -Ⅰ	H: +Ⅰ	C: -Ⅱ
i) CHF3		F: -Ⅰ	H: +Ⅰ	C: +Ⅱ
j) CH4O		O: -Ⅱ	H: +Ⅰ	C: -Ⅱ
k) N2O4		N: +Ⅳ	O: -Ⅱ
l) NO		N: +Ⅱ	O: -Ⅱ
m) HNO3		H: +Ⅰ	N: +Ⅴ	O: -Ⅱ
n) H3O+		H: +Ⅰ	O:-II
o) NH+4		N: -Ⅲ	H: + I

Solution:

The most common ions would be Br^–, O^2-, K⁺ , Al³⁺ , Ca²⁺ , N^3-

	K+	Al3+	Ca2+
Br–	KBr	AlBr3	CaBr2
O2-	K2O	Al2O3	CaO
N3-	K3N	AlN	Ca3N2

Soluzione:

Applicazione:
— 1·NaCl ⇄ 1·Na⁺ + 1·Cl^–
— c(NaCl) = c(Na⁺) = c(Cl^–)
— L_p(NaCl) = c(Na⁺)·c(Cl^–)
— c = n/V = m/M/V
— c(NaCl) = 100g/58.5 g/mol / 0.4 l = 4.27 mol/l
— L_p(NaCl) = 4.27 mol/l · 4.27 mol/l = 18.26 mol²/l²

Solution:

1. · Réactions impliquées:
2.     Cu²⁺ + 2·e^– → Cu, z = 2
3.     6·H₂O → 4·H₃O⁺ + O₂ + 4·e^– ; z = 4
4. · m/M = I·t/(z·F)
5.     t = F·m·z/(I·M)
6.     t = 96485·0.8·2/(0.18·63.5)
7.     t = 13506 s = environ 4 h
8. · m/M = n(O₂) = I·t/(z·F)
9.     n(O₂) = 0.18·13506/(4·96485) = 0.00629 mol
10. · p·V = n·R·T
11.     V(O₂)= n·R·T/p
12.     V(O₂) = 0.00629·8.314·298/101300
13.     V(O₂) = 0.000154 m³ = env. 154 ml
14. · Unités: [mol·J/(mol·K) ·K / N/m²]

Soluzione:

Applicazione:
— CaSO₄ ⇄ Ca²⁺ + SO2-4      L_p(CaSO₄) = c(Ca²⁺)·c(SO2-4)
— Ionenprodukt(CaSO₄) = 10^-2 mol/l · 10^-4 mol/l = 10^-6 mol²/l²
— Quindi: 10^-6 mol²/l² < L_p (=2·10^-5 mol²/l²).
— Quindi la soluzione è insatiabile.

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