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October 16, 2026 at 4:42 am in reply to:
Conjugés, divers exemples 1 (SB-12)
S'il vous plaît, complétez la table suivante, résolvant le premier exemple.
Devrait être il pour une substance utilisée z.B. Aucun conjugaux base donner (z.B. F^-), alors devrait le contenu de la cellule est marqué par un “X”.
à base acide conique
H₂O OH^– H₃O⁺
HS^–
F^–
H₂PO–4
PO-34
H₃PO4
NH₃
NO–2
#5599
Solution:
à base acide conique
H₂O OH^– H₃O⁺
HS^– S^2- H₂S
F^– x HF
H₂PO–4 HPO2-4 H₃PO₄
PO-34 x HPO2-4
H₃PO₄ H₂PO–4 x
NH₃ NH–2 NH+4
NO–2 x HNO₂
October 16, 2026 at 4:42 am in reply to:
Redox, electrolysis, 3 (Redox-53)
15 m² a metal surface area should with a 35 µm thick chromic layer coated are. Chromes has a density of 7.15 g/cm³.
(a) Calculate the mass of chromium to be excreted.
(b) Separation shall be made from an CrO₃ – solution. How long to be electrolyzed if the current strength is 450 A And the effectiveness is 90%.
Note : identical exercise also at the same time, subjects salts.
#9529
Solution:
Note: 35 µm correspond to 35·10^-6m
1. The volume of the chromium layer to be added shall be:
  V= O·h = 15m^2·35·10^-6m = 0.000525 m³ = 525 cm³
2. With density = m/V:
  m=density·V= 7.15g/cm³ · 525 cm³ = 3753.75 grams
3. CrO₃, participants ions: 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 (at 100% efficiency)
5. Efficiency of 90%: 103164 seconds
October 16, 2026 at 4:42 am in reply to:
Entalpia di reazione calcolare 3 (thermo-15)
Il glucosio (C₆H₁₂O₆) può essere trasformato in diverse forme: aerobico (con O₂) e anaerobico (senza O₂).
Formulare per entrambe le reazioni le equazione di reazione e calcola ciascuna la entalpia di reazione standard.
Nota I: se i ΔHof sono sconosciuti, si calcola generalmente con x, y, ecc.
Nota II: indicare stato di aggregazione
#7565
Soluzione:
ΔHof(C₆H₁₂O₆) = -1268 kJ/mol
ΔHof(O₂)(g) = 0 kJ/mol
ΔHof(C₂H₅OH)(l) = -278 kJ/mol
ΔHof(H₂O)(g) = -286 kJ/mol
ΔHof(CO₂) = -393 kJ/mol
a) Reazione di combustione:
C₆H₁₂O₆ + 6·O₂(g) ⇄ 6·CO₂(g) + 6·H₂O(l)
ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
ΔHoR = (6·-393 + 6·-286)-(-1268+6·0) = – 2806 kJ/mol
b) Fermentazione:
C₆H₁₂O₆ ⇄ 2·C₂H₅OH(l) + 2·CO₂(g)
Nota: C₂H₅OH si chiama alcol
ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
ΔHoR = (2·-278 + 2·-393)-(-1268) = – 74 kJ/mol
October 16, 2026 at 4:42 am in reply to:
Redox, electrolysis, 2 (Redox-52)
A silver nitrate solution is electrolyzed at a current strength of 2.54 A. The mass of silver is released in 45 minutes. The efficiency is 90%.
Note : identical exercise also at the same time, subjects salts.
#9527
Solution:
Silver nitrate: AgNO₃, metallion: Ag⁺
Note 1: m/M = I·t/(z·F)
Note 2: Assume 100% efficacy in a first step.
From m/M = I·t/(z·F) follows:
m = (I·t·M)/(z·F) = (2.54·45·60·107.87) / (1·96485) = 7.67 g
Efficacy of 90%: 6.9 grams
Units: [(A·s·g/mol)/(C/mol)], [A] = [C/s]
October 16, 2026 at 4:42 am in reply to:
Konjugiert_2 (SB-11)
Désigner suivantes acides-bases-les couples plus précisément ('x est conjugaux acide / base de y [traduction manquante]'). En particulier est valable il donc, à le exercice a) ou b) chaque tous quatre components avec l'autre dans la relation à mettre en place.
a) H₂CO₃ + 2 H₂O ⇄ CO2-3 + 2 H₃O⁺
b) NH₃ + H₂O ⇄ NH+4 + OH^–
#5597
Solution:
1. a) le H₂CO₃ est le conjugé de acide à CO2-3
     CO2-3 est la conjugaison de base à H₂CO₃
     H₂O est la base conjugée à H₃O⁺
     H₃O⁺ est la conjugaison de acide à H₂O
2. b) NH₃ est la base conjuguée à NH+4
     NH+4 est l'acide conjugé à NH₃
     H₂O est la conjugaison de acide à OH^–
     OH^– est la base conjuguée à H₂O
October 16, 2026 at 4:42 am in reply to:
Entalpia di reazione calcolare 2 (thermo-14)
Quando si vendono 60 grammi di zucchero di uva, vengono rilasciati x kJ. Calcola il ΔHof.
#7563
Soluzione:
M(Traubenzucker) = M(C₆H₁₂O₆) = 180 grammi
Con 60 g C₆H₁₂O₆ x kJ, a un Mol (=180 g) quindi 3·x.
ΔHof(C₆H₁₂O₆) = 3·x kJ/mol.
October 16, 2026 at 4:42 am in reply to:
Konjugiert_1 (SB-10)
Qu'est-ce que les couples acide-base conjugués?
#5595
Solution:
Donne une acide sein(e) protons à partir de, ainsi se produit le parties de base (base parce que il oui théoriquement le proton encore une fois enregistrez pourrait). Ces parmi ceux-ci, les couples sont couples acide-base conjugués nommé.
Exemple: H₃O⁺ comme acide est à H₂O. Donc serait H₃O⁺ / H₂O le acides-bases-un couple. Trivial but: note, C'est alors que l'acide (H₃O⁺) apparaît en premier.
October 16, 2026 at 4:42 am in reply to:
Entalpia di reazione calcolare 1 (thermo-13)
Calcola le [traduzione mancante]-entalpia di reazione seguenti reazione:
H₂(g) + 1/8 S₈ (s) ⇄ H₂S(g)
#7561
Soluzione:
I seguenti valori per ΔHof possono essere tratti da tabelle. Se non si trova un valore: generalmente proseguire con x, y, ecc.
H₂(g) + 1/8 S₈ (s) ⇄ H₂S(g)
ΔHof(H₂(s)) = 0 kJ/mol
ΔHof(S₈(s)) = 0 kJ/mol
ΔHof(H₂(g)) = x kJ/mol
ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
ΔHoR = ΔHof(H₂S(s)) – ( ΔHof(H₂(s)) + 1/8 ΔHof(S₈(s)) )
ΔHoR = x – (0 + 0) = x
October 16, 2026 at 4:42 am in reply to:
Redox, electrolysis, 1 (Redox-51)
Copper is obtained electrolytically from Cu²⁺ – ions. In a electrolytic cells is a Cu²⁺ -withdrawn solution 30 min with a power intensity of I = 2.5 A electrolytic. Calculate the separated mass m(Cu) at an electrolysis efficiency of 92.5%.
Note 1: m/M = I·t/(z·F)
Note 2: Assume 100% efficacy in a first step.
Note : identical exercise also at the same time, subjects salts.
#9525
Solution:
From m/M = I·t/(z·F) follows:
m = (I·t·M)/(z·F) = (2.5·30·60·63.546) / (2·96485) = 1.48 g
Efficiency of 92.5%: 1.37 grams
Units: [(A·s·g/mol)/(C/mol)], [A] = [C/s]
October 16, 2026 at 4:42 am in reply to:
Compare l’entropia 2 (thermo-12)
Diciamo la molecola C₂H6 e CH₄. quale di queste due associazioni ha una maggiore entropia.
#7559
Soluzione:
Nota: C₂H₆ ha più atomi e quindi una maggiore entropia dà il risultato corretto, ma la motivazione è completamente sbagliata! C₂H₆ ha quindi un'entropia maggiore di CH₄ perché il C₂H₆ offre altre possibilità aggiuntive di disposizione a causa del suo legame C-C: il C₂H₆ può effettuare rotazioni interne lungo il suo asse C-C, che contribuiscono anche al valore dell'entropia totale.
October 16, 2026 at 4:42 am
in reply to:
Le comportement d’un acide ou d’une base (SB-03)
Formulez des réactions pour les substances données suivantes. L'abréviation LM désigne le solvant.
Il ne s'agit pas d'évaluer si Les requises réactions possible sont ou pas, il s'agit c'est pourquoi, que z.B. À une requises réaction acide le accordée particules un H⁺ dégage et le les solutions cette alors prend en charge.
a) HBr, réaction acide, LM: eau
b) HF, réaction de base, LM: eau
(c) HF, réaction acide, LM: eau
d) HF, réaction acide, LM: NH₃
e) H₂O, réaction acide, LM: eau
f) CH₃COOH, réaction acide, LM: eau
g) CH₃OH, réaction de base, LM: NH₃
h) CH₃OH, réaction acide, LM: eau
#5593
Solution:
1. a) HBr + H₂O ⇄ Br^– + H₃O⁺
2. b) HF + H₂O ⇄ FH+2 + OH^–
3. c) HF + H₂O ⇄ F^– + H₃O⁺
4. d) HF + NH₃ ⇄ F^– + NH+4
5. e) H₂O + H₂O ⇄ OH^– + H₃O⁺
6. f) CH₃COOH + H₂O ⇄ CH₃COO^– + H₃O⁺
7. g) CH₃OH + NH₃ ⇄ CH₃OH+2 + NH–2
8. h) CH₃OH + H₂O ⇄ CH₃O^– + H₃O⁺
October 16, 2026 at 4:42 am in reply to:
Electrolysis – a redox reaction, cloze exercise (Redox-50)
Complete the following gaps with meaningful content. The reaction under consideration is the electrolysis of water.

… charged oxonium ions … … travel in the electric field to the … charged electrode (cathode), where they each receive an electron … Hydrogen atoms are formed, which combine with another H atom created by… to form a hydrogen molecule. What's left are water molecules.

The isolated gaseous hydrogen rises at the cathode, where the cathode space becomes … the negatively charged hydroxide anions migrate to the … anode, where hydroxidions with protons (H⁺ and H₃O⁺) neutralize to water or convert to oxygen at the anode under …

Here, too, the isolated oxygen rises as a gas at the anode, at the same time the anode space becomes…. The protons created travel in the direction of… – analogous to the processes in the cathode space.
#9523
Solution:
Positive loaded [translation unavailable]-ions (H₃O⁺ ) traveling in the electrical field to the negative loaded electrodes (catheter), where they each a electron to take up. . In doing so to be created hydrogen-atoms, the itself with a other, through reduction of which: H-atom to a hydrogen molecule to unite. What's left are water molecules.
The isolated gaseous hydrogen rises at the cathode, where the cathode space becomes basic . The negatively charged hydroxide anions migrate to the positive anode, where negative hydroxidions neutralize into water with protons or convert into oxygen at the anode under electron emission .
Also here increases the separated oxygen as [translation unavailable] at the [translation unavailable] on, at the same time is the anodic space acidic . The of which: protons traveling in direction catheter – by analogy to the measures to be taken in the cathode room.
October 16, 2026 at 4:42 am in reply to:
Différence entre H et proton (SB-02)
Est-ce qu'un H⁺ est le même qu'un proton ?
#5591
Solution:
Dans le principe oui, ABER le respect: il est le proton d’un à nu les atomes d'hydrogène ('H sans électron') désigné. Par mémoire: les éléments sont composés des protons et éventuellement des neutrons (ainsi qu'un certain nombre d'électrons). Ainsi, tous les éléments ont aussi un certain nombre de protons, l'hélium par exemple, deux protons. Ces protons faire mais NICHT le caractéristiques une acide de.
October 16, 2026 at 4:42 am in reply to:
Redox, car battery, sales (Redox-42)
A car batteries be full loaded and containing 1.5 litres a 30% mixtures of sulphuric acid. '[translation unavailable]' be the concentration only still 20%. How much lead or lead oxide was applied each time?
Note: the density the 30% sulphuric acid-solution of which: 1.84 g/cm³. The density is equal to the discharge.
#9521
Solution:
1. % refers to mass percentages
2. 1 litres has a mass of 1840 grams. 1.5 litres thus 2760 g
3. In total are 10% implemented (of 30% → 20%), i.e. 276 grams
4. The balanced reaction equation is:
  1·Pb + 1·PbO₂ + 2·H₂SO₄ ⇄ 2·PbSO₄ + 2·H₂O
This leads to the following scenario:
substance M(g/mol) m(g) n(mol)
H₂SO₄ 98 276 2.816
Pb 207.2 291.77 1.408
PbO₂ 239.2 336.79 1.408
Es have been thus [translation unavailable] 291.77 grams lead and 336.79 grams lead oxide (PbO₂) implemented
October 16, 2026 at 4:42 am
in reply to:
Entalpie di reazione o tropi di reazione (thermo-11)
Riflettere a te per ogni il seguenti procedimenti, se ciascuno una aumento o una riduzione il entropia e il entalpia del prodotti chimici sistemi a aspettare è (z.B. ΔHoR [traduzione mancante]0 [traduzione mancante].):
CO₂(s) sublimato a CO₂(g)
riscaldamento di CaCO₃(s)
CO₂(g) + CaO(s) → CaCO₃(s)
H₂O(s) → H₂O(l) → H₂O(g) → 0.5·O₂(g) + H₂(g)
Ba(OH)₂(s) + 2·NH₄SCN(s) → Ba(SCN)₂ + 2·NH₃(g) + 2·H₂O(l)
NH₃(g) + HCl(g) → NH₄Cl(s)
#7557
Soluzione:
1. CO₂(s) sublimato a CO₂(g):
     ΔSoR [traduzione mancante]0, ΔHoR [traduzione mancante]0
     ΔSoR > 0, poiché CO₂ viene trasferito da solido (s) a gas (g)
     ΔHoR > 0, dal momento che il CO₂ viene prima liquefatto e poi trasferito nello stato gassoso, processo endotermico
2. riscaldamento di CaCO₃(s):
     ΔSoR [traduzione mancante]0, ΔHoR [traduzione mancante]0
     ΔSoR in caso di aumento della temperatura, le particelle si muovono più intensamente (ora e indietro), dS quindi > 0
     ΔHoR > 0, dal momento che il riscaldamento, cioè l'energia deve essere investita, processo termico
3. CO₂(g) + CaO(s) → CaCO₃(s):
     ΔSoR [traduzione mancante]0, ΔHoR [traduzione mancante]0
     ΔSoR < 0, in quanto due particelle diventano una particella.
     Esperimento (o potrebbe essere calcolato), processo esotermo, ΔHoR < 0
4. H₂O(s) → H₂O(l) → H₂O(g) → 0.5·O₂(g) + H₂(g):
     ΔSoR [traduzione mancante]0, ΔHoR [traduzione mancante]0
     ΔSoR > 0, dal momento che i passaggi da fisso → liquido → gassoso, inoltre si formano più particelle
     ΔHoR > 0, esperienza quotidiana: trasferimento da solido → liquido ecc. necessita di energia, processo endotermico
5. Ba(OH)₂(s) + 2·NH₄SCN(s) → Ba(SCN)₂(aq) + 2·NH₃(g) + 2·H₂O(l):
     ΔSoR [traduzione mancante]0, ΔHoR [traduzione mancante]0
     ΔSoR [traduzione mancante]0, lì da i due fabbricanti una liquidità, un [traduzione mancante] e nel acqua acquatizzati Ba(SCN)₂ presente
     L'esperimento, una tipica miscelazione a freddo. Quando si mescolano i due sali, la miscela si raffredda, processo termico, ΔHoR > 0
6. NH₃(g) + HCl(g) → NH₄Cl(s):
     ΔSoR [traduzione mancante]0, ΔHoR [traduzione mancante]0
     ΔSoR < 0, di due particelle (gas) diventa una particella
     ΔHoR < 0, ma giustificabile solo con un calcolo
October 16, 2026 at 4:42 am in reply to:
Définition de l’acide (SB-01)
Qu'est-ce qu'un acide ou une base?
#5589
Solution:
Il existe différentes définitions. Les le plus simple: une acide est une connexion, quelles H⁺ (aussi protons nommé) déposer peut, z.B. HF. Une base est donc une substance qui peut absorber H⁺.
October 16, 2026 at 4:42 am in reply to:
Redox, amp hours, Ah (Redox-41)
In the case of a car battery, the following is added: 100 Ah, 12 V
So how many electrons (in mol) can flow in 5 hours?
#9519
Solution:
1. Note: the number the flowing electrons is independently of the time, it [translation unavailable] thus not on it at, whether or not the total loading in 5 h or also only in a hours flowing
2. Note: the additionally A (amplifiers) is a indication, how many electrons per [translation unavailable] flow can: 1 C/s = 1 amplifiers
It holds general: a battery with 100 [translation unavailable] can z.B. 10 hours with 10 amplifiers operated are, or 1 hour with 100 ampere etc. Not to be confused or overlooked: ampere is the indication of a charge per second (more accurately: 1 coulomb per second)
In principle, the Ah indicator is nothing more than a hidden Coulomb or charge indicator:
1 C = 1 A·s = 1 A·h/3600 = 0.000277 [translation unavailable]
100 Ah is therefore 360000 Coulombs
The charge of an electron is 1.602·10^-19 C. So a Coulomb is equal to 6.242·10¹⁸ electrons (per second !)
In total are thus 2.24·10²⁴ electrons = approx. 3.73 mol electrons
October 16, 2026 at 4:42 am in reply to:
Entalpia di reazione calcolare (thermo-10)
Come si calcola l'entalpia di reazione di una qualsiasi reazione?
#7555
Soluzione:
Risposta breve, molto grosso: “Prodotto” – “Educato”
Più dettagliato e più preciso: la somma delle entalpi di reazione (standard) dei prodotti meno il Somma delle entalpi di reazione (standard) degli reagenti.
ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
October 16, 2026 at 4:42 am in reply to:
La dépendance à la température de la constante d’équilibre K (thermo-42)
Calculer ΔGoR pour les températures 298K, 900K et 2300K de la réaction de combustion de N₂ à NO. Les températures données de 900 K correspondent à la température de l'échappement et de la combustion (2300K),
Remarque: calculez la réaction pour 1 particules (ou 1 mol) NO.
#5587
Solution:
L'équation de réaction suivante est examinée: ½·O₂(g) + ½·N₂(g) ⇄ NO (g)
Tableau des valeurs:
ΔHof(kJ·mol^-1) S^o(J·K^-1·mol^-1)
O₂ (g) 0 205.03
N₂ (g) 0 191.61
NOO (g) 90.37 210.62
1. équation de réaction:
  ½·O₂(g) + ½·N₂(g) ⇄ 1·NO (g)
2. ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
  ΔHoR = (1· ΔHof(NO,g)) – (½· ΔHof(O₂, g) + ½· ΔHof(N₂,g))
  ΔHoR = (90.37) – (½·0 + ½·0) = +90.37 kJ/mol
3. ΔSoR= S^o(Produkte) – S^o(Edukte)
  ΔSoR = (1·S^o(NO,g)) – (½·S^o(O₂, g) + ½·S^o(N₂,g))
  ΔSoR = (210.62)-(½·205.03 + ½·191.61) = +12.3 J·mol^-1·K^-1
4. ΔGoR = ΔHoR – T· ΔSoR
  ΔGoR(T=298K) = 90.37 – (298)·0.0123 = 86.7 kJ/mol
  ΔGoR(T=900K) = 90.37 – (900)·0.0123 = 79.3 kJ/mol
  ΔGoR(T=2300K) = 90.37 – (2300)·0.0123 = 62.08 kJ/mol
5. avec dG = -RTln K suit: K = e^(-dG/R·T), R=0.008314 kJ/mol·K
  ΔGoR(T=298K) = 86.7 kJ/mol → K= 6.34·10^-16
  ΔGoR(T=900K) = 79.3 kJ/mol → K= 2.5·10^-5
  ΔGoR(T=2300K) = 62.08 kJ/mol → K= 0.03
Discussion: il action se pour une [traduction manquante] réaction ( ΔHoR > 0). Pour stimuler la réaction Il serait donc nécessaire d'augmenter la température. Ceci est confirmé par le calcul de la constante d'équilibre K: plus la température est élevée, plus la valeur de K elle-même devient grande. Es montre se mais, que le équilibre mais aussi à à hauteur températures toujours encore relativement fortement sur le la page le réactifs est situé, un utilisation d’un des catalyseurs offre se donc à.
October 16, 2026 at 4:42 am in reply to:
Joule, potere calorifico inferiore, alcalino (thermo-07)
Questo è un esperimento: un alcalino scalda l'acqua in un vaso di vetro. Indicare un'equazione in cui viene fornito l'equivalente 'memorizzato' nella noce potere calorifico superiore.
#7553
Soluzione:
La noce di avena brucia (non completa), riscalda l'acqua, il vetro e l'ambiente.
1. La noce di avena brucia (non completa), riscalda l'acqua, il vetro e l'ambiente.
2. La noce non brucia completamente, solo la massa della noce bruciata fornisce energia, quindi può essere considerata solo Δm(Nuss)
3. All'inizio tutto ha la temperatura T₀
4. W si riferisce all'acqua, G al vetro, L all'aria
5. Dopo il riscaldamento l'acqua ha la temperatura T_W, il vetro T_G, l'aria T_L
6. H·Δm(Nuss) = c_w·m_w·(T_W-T₀) + c_G·m_G·(T_G-T₀)+c_L·m_L·(T_L-T₀) + radiamento termico
October 16, 2026 at 4:42 am
in reply to:
Redox, car battery (Redox-40)
In the case of an automotive battery, explain the following processes:
Unloading
Shopping
#9517
Solution:
Not done yet
October 16, 2026 at 4:42 am in reply to:
Calculer la température de séchage (thermo-41)
Calculez les point d’ébullition du les eaux !
#5585
Solution:
Il s'applique: H₂O(l) ⇄ H₂O(g).
Pour cet équilibre chimique: K = c(H₂O(g))/c(H₂O(l)) = 1
Avec dG = -R·T·ln(K) suivie par K = 1: dG = 0
En outre: dG = dH-T·dS (=0)
donc T = dH/dS
1. T = dH/dS
     ΔHoR = ΔHof(H₂O, g) – ΔHof(H₂O, l)
     ΔHoR = -241.82 – (-285.83) = 44.01 kJ/mol
     ΔSoR = S^o(H₂O, g) – S^o(H₂O, l)
     ΔSoR = 188.85 – 69.91 = 118.92 J/(mol·K) = 0.11892 kJ(mol·K)
2. T = dH/dS = 44.01 / 0.11892 = 370 K ou Env. 97℃
Les calculé (!) température de séchage du les eaux est donc env. 97℃ Les déviation pour le habituels la valeur de 100℃ il vient par conséquent, parce que dH ou dS en fait aussi (peu) de le température dépendre.
October 16, 2026 at 4:42 am in reply to:
Joule, potere calorifico inferiore (thermo-06)
Determina la potere calorifico inferiore di brennsprit ('Spiritus') dalle seguenti indicazioni: 500 grammi di acqua vengono riscaldati da 1 grammi di brennsprit da 20.4℃ a 22.7℃
Nota: il potere calorifico inferiore H è in [traduzione mancante] (oppure kJ) per kg indicato
#7551
Soluzione:
1. L'abbreviazione S è Spiritus
2. Anche dalla considerazione delle unità (il lato destro dell'equazione in joule, quindi anche il lato sinistro deve essere in J)
3. H·m_S = m_w·c_w·ΔT
4. H = m_w·c_w·ΔT / m_s = 0.5 kg · 4182 J/kgK · 2.3 K / 0.001 kg = 4'809300 J/kg = 4809 kJ/kg
October 16, 2026 at 4:42 am in reply to:
Redox, Redox table, Nernst 4 (Redox-33)
(a) What voltage is calculated for the Daniell element if at 298 K the concentration of the zinc sulphate solution and that of the copper sulphate solution is 1 mol/L each. Please specify the generally applicable equation for this.
(b) Calculate the new voltages assuming that after every 15 minutes the concentrations will decrease by 0.1 mol/L We have changed.
#9515
Solution:
From the previous exercise can be taken E₁ and E₂. Also the presentation is simplified a little because the concentration of the solid is 1 mol/l
1. (a) It applies from earlier:
     · E₁(Zn/Zn²⁺)=-0.76 + 0.059/2·log(c(Zn²⁺))
     · E₂(Cu/Cu²⁺)= 0.35 + 0.059/2·log(c(Cu²⁺))
     · ΔE = E₂ – E₁
     = 0.35 + 0.059/2·log(c(Cu²⁺)) – (-0.76 + 0.059/2·log(c(Zn²⁺)))
     = 1.11 + 0.059/2 · ( log(c(Cu²⁺)) -log(c(Zn²⁺)) )
     = 1.11 + 0.059/2 · log(c(Cu²⁺)/c(Zn²⁺))
(b) The concentration of copper ion increases, the concentration of zinc increases.
ΔE = 1.11 + 0.059/2 · log(c(Cu²⁺)/c(Zn²⁺))
c(Cu²⁺) c(Zn²⁺) ΔE
1 1.0 1.11
0.9 1.1 1.107
0.8 1.2 1.105
0.6 1.4 1.099
0.4 1.6 1.092
0.2 1.8 1.082
0.001 1.999 1.012
October 16, 2026 at 4:42 am in reply to:
Joule, miscela, 1 (thermo-05)
Si mescolano 1 kg di ghiaccio in scioglimento e 1 kg di ghiaccio seduto. Qual è la temperatura della miscela quando tutto il ghiaccio si scioglie?
#7549
Soluzione:
1. L'acqua calda si raffredda, l'acqua fredda si riscalda
2. Le abbreviazioni seguenti si applicano: H per caldo, K per freddo, M per miscela
3. Il ghiaccio deve essere fonduto prima (L_f è l'energia necessaria) e poi può essere riscaldato fino alla temperatura di miscelazione
4. Q(100℃ → T_M) = Q(Eis schmelzen, 0℃ → T_M)
5. m_H · c_w · (100℃ – T_M) = m_K·L_f + m_K · c_w · (T_M – 0℃ )
6. Convertire il Celsius in Kelvin, m_H = m_K = m
7. m·c_w·(373K-T_M) = m·L_f +m·c_w·(T_M-273K)
8. T_M = [c_w·(373k+273k)-L_f]/(2·c_w)
9. T_M = [4.185kJ/kgK·(373k+273k)-335.5kJ]/(2·4.185kJ/kgK) = 283.2 K = 10.2℃
October 16, 2026 at 4:42 am in reply to:
Énergie libre de Gibbs de réaction, eau à l’intérieur Péroxyde d’hydrogène (thermo-40)
Es réagir H₂(g) avec O₂(g). Les évaluations avec l'aide le à titre d'exemplaire enthalpie de réaction ou [traduction manquante] H₂O(g) ou H₂O₂(g) formé est. La température de réaction est de 800 Kelvin.
#5583
Solution:
Es laissez se deux réactions formuler:
a) H₂(g) + 1/2·O₂(g) → H₂O(g) et
b) H₂(g) + O₂(g) → H₂O₂(g)
Le respect: afin que les résultats avec l'autre comparé sont peuvent doit on se sur les même nombre éduqué H₂O ou H₂O₂ en rapport avec!
Table des valeurs indiquant que toutes les particules impliquées sont en état gazeux:
ΔHof(kJ·mol^-1) S^o(J·K^-1·mol^-1)
H₂ (g) 0 131
O₂ (g) 0 205
H₂O (g) -242 189
H₂O₂ (g) -136 233
1. Réaction a) produit H₂O
     Définition: ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
     ΔHoR(H₂O) = (-242) – (0 + 1/2·0) = -242 kJ/mol
     ΔSoR(H₂O) = (189) – (131 + 1/2·205) = -44.5 J/(mol·K)
     Définition: ΔGoR = ΔHoR – T· ΔSoR
     ΔGoR(H₂O) = -242 – 800·-0.0445 = -206.4 kJ/mol
2. Réaction b) produit H₂O₂
     Définition: ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
     ΔHoR(H₂O₂) = (-136) – (0 + 1/2·0) = -136 kJ/mol
     ΔSoR(H₂O₂) = (233) – (131 + 205) = -103 J/(mol·K)
     Définition: ΔGoR = ΔHoR – T· ΔSoR
     ΔGoR(H₂O₂) = -136 – 800·-0.103 = -53.6 kJ/mol
Le calcul montre que bien que les deux réactions soient exergonées ( ΔGoR < 0) et doivent donc se dérouler volontairement, la réaction a) est presque un facteur 4 plus exergonée et se déroulera donc de préférence.
October 16, 2026 at 4:42 am in reply to:
Redox, Redox table, Nernst 3 (Redox-32)
What voltage is calculated for the Daniell element if at 298 K the concentration of the zinc sulphate solution is 0.001 mol/L and that of the copper sulphate solution is 0.8 mol/L.
#9513
Solution:
1. Observation: Zn → Zn²⁺
     · c(reduzierte Form) = c(Zn) = 1.0 (solids)
     · c(oxidierte Form) = c(Zn²⁺)
     · E₁(Zn/Zn²⁺)=-0.76 + 0.059/2·log(c(Zn²⁺)/c(Zn))
     · E₁ = -076 + 0.059/2·log(0.001/1) = -0.849 V
2. Monitoring and monitoring Cu²⁺ → Cu
     · c(reduzierte Form) = c(Cu) (solids)
     · c(oxidierte Form) = c(Cu²⁺)
     · E₂(Cu/Cu²⁺)=0.35 + 0.059/2·log(c(Cu²⁺)/c(Cu))
     · E₂ = 0.35 + 0.059/2·log(0.8/1.0) = +0.347 V
3. ΔE = E_{Kathodenreaktion} – E_{Anodenreaktion}
     · The cathode reaction is defined as a reduction
     · Reduction: Cu²⁺ → Cu, which is E₂
     · ΔE = E₂ – E₁ = 0.347-(-0.849)=1.196 V
October 16, 2026 at 4:42 am in reply to:
Enthalpie de réaction Différence de masse 2 (thermo-31)
Sur wikipédia (https://de.wikipedia.org/w/index.php?title=%C3%84quivalenz_von_Masse_und_Energie&oldid=245822194) est à partir de applications pour les conversion de masse à l'égard de Einstein, que [traduction manquante] réchauffement pour 50 le degré célsius d’un mètre cube eau (adoption par le Conseil: masse 1 [traduction manquante]) se une augmentation de la masse pour 2.3 microgramme déposer.
Confirmez cette valeur avec une facture.
#5581
Solution:
Pour calculer l'augmentation de masse, nous devons suivre les étapes suivantes:
→ calcul de l'énergie thermique fournie
→ Calculer l'augmentation de masse selon Einstein
1. Q = m·c_w·ΔT
2. Q = 1000kg · 4,18 kJ/kg · 50 ℃
3. Q = 2.09·10⁸ J
4. ΔE=Δm·c² ou Δm = ΔE/c²
5. Δm = 2.09·10⁸ [kg·m²/s²]/(300'000'000 m/s)² = 2.32·10^-9 kg
October 16, 2026 at 4:42 am
in reply to:
Redox, Redox table, Nernst 2 (Redox-31)
Calculate the potential for the following given solutions or reactions.
E = E^o + 0.059/n · log(c^a(Oxidierte Form)/ c^b(Reduzierte Form))
Zn → Zn²⁺, c(Zn²⁺) = 0.1 mol/L
2 Cl^– → Cl₂, c(Cl^–) = 2 mol/L
Mn²⁺ + 12 H₂O → MnO–4 + 8 H₃O⁺ , c(Mn²⁺) = x, c(MnO–4) = y
#9511
Solution:
1. Zn → Zn²⁺, n= 2 , E^o(Zn/Zn²⁺) = -0.76 V
     · c(Reduzierte Spezies) = c(Zn) = 1, there more solid substance
     · c(Oxidierte Spezies) = c(Zn²⁺)= 0.1 mol/L
     · E = -0.76 + 0.059/2·log(0.1/1) = -0.7895 V
2. 2 Cl^– → Cl₂, n=2, E^o(2 Cl^–/Cl₂) = 1.36 V
     · c(Reduzierte Spezies) = c(Cl^–) = 2 mol/L
     · c(Oxidierte Spezies) = c(Cl₂)= 1, there [translation unavailable] substance
     · E = 1.36 + 0.059/2·log(1/c²(Cl^–) )
     · E = 1.36 + 0.059/2·log(1/4) = 1.34 V
3. Mn²⁺ + 12 H₂O → MnO–4 + 8 H₃O⁺ , E^o(…) = 1.51 V
     · The ox numbers at H₂O and H₃O⁺ do not change !
     · Ox numbers: Mn: +Ⅱ → +Ⅶ
     · c(Reduzierte Spezies) = c(Mn²⁺) = x
     · c(Oxidierte Spezies) = c(MnO^-4) = y
     · E = 1.51 + 0.059/5·log(y/x)
October 16, 2026 at 4:42 am
in reply to:
Joule, applicazioni 3 (thermo-04)
Quanti kilojouli di calore si generano quando un'auto di 1000 kg viene completamente frenata a 95 km/h?
Quanto kg acqua potrebbe si di 10℃ su 60℃ con il quantità di energia il precedenti esercizio riscaldamento?
Una goccia d'acqua cade da una nuvola e, teoricamente, si riscalderà a quante gradi Celsius a causa della conversione di energia.
#7547
Soluzione:
1. E_kin = ½ · m · v² = ½ · 1000kg · (26.38m/s) ^2 = 348'187 J
2. E_kin = Q = c_w · m · ΔT
  m = Q/(c_w·ΔT) = 348'187 J / (4182 J·kg^-1·K^-1 · 50 K) = 1.665 kg
3. E_pot = Q
     m·g·h = c_w · m · ΔT
     Le masse si riducono, presumendo h = 5000 m
     ΔT = g·h / c_w = 10m/s² · 5000 m / 4182J·kg^-1·K^-1 = 11.7℃
October 16, 2026 at 4:42 am in reply to:
Enthalpie de réaction, différence de masse 1 (thermo-30)
Dans le cas de la réaction de gaz à effet de serre 2H + O → H₂O, l'énergie de réaction de 571.6 kJ/mol est libérée.
Remarque: [J] = [kg m² s^-2]
#5579
Solution:
1. À l'égard de Einstein est valable E=m·c² ou ΔE=Δm·c²
2. c est la vitesse de la lumière, 300'000 km/s
3. Réformer le Einstein'- à l'intérieur l'équation après m ou Δm
4. Δm = ΔE/c²
5. Δm = 571.6·1000[kg·m²/s²]/(300'000'000 m/s)² = 6.35·10^-12 kg
Les 6.35·10^-12 kg calculés se rapportent à un Mol. Maintenant, si l'on veut remonter à une seule molécule d'eau, on doit encore la diviser par le nombre d'Avogadro (6.022·10²³) et on obtient la diminution théorique de la masse par molécule:
6.35·10^-12 kg / 6.022·10²³ = 1.05·10^-35 kg
ABER: à l'égard de 'bonne question.Réseau', cette fois-ci mais une utilisables réponse:
Les accroissement à masse par est pas par le plus simple réservation de énergie, il s'agit par les DIREKTE la conversion de énergie à matériel (les énergie KONDENSIERT QUASI ZU MATERIE) ceci se produit cependant, à ainsi à hauteur températures, que les classiques la physique pas plus est valable. Avec une alimentation pure d'énergie, il n'y a qu'une augmentation de la température.
October 16, 2026 at 4:42 am
in reply to:
Redox, Redox table, Nernst 1 (Redox-30)
What 's the Nernst equation ?
What is the Nernst equation for diluted solutions of the same metal
A concentrated Cu²⁺ – solution and a diluted Cu²⁺ – solution are given, each containing a Cu electrode. Solutions, only separated by a diaphragm, connected to each other? Use of which: other the concept reduction, oxidation, court of Justice at the catheter or [translation unavailable].
How do you calculate the voltage between two different half-cells ?
Calculate the potential of a silver cell with a concentration of 0.1 mol/l
Calculate the potential of a silver cell with a concentration of 0.01 mol/l
What is the voltage between two half-cells with a voltage of 0.1 and 0.01 mol/L?
A zinc half cell is diluted by a factor of 1000. Calculate the voltage drop and the new potential.
#9509
Solution:
What happens if the
1. E = E^o + 0.059/n · log(c^a(Oxidierte [translation unavailable])/ c^b(Reduzierte [translation unavailable])
2. E = E^o + 0.059/n · log(c(verdünnte Lösung)/c(konzentrierte Lösung))
3. Finally, the two solutions must have the same concentration.
  · The concentration must decrease at the higher concentration solution. Would be, that the copper ions to elementary copper are (Cu²⁺ → Cu), it must thus a reduction to take place, according to the definition at the catheter.
  · Conversely, the deeper concentration must be increased. This goes only so, that fixed copper (from the electrodes) gained is, Cu → Cu²⁺, (oxidation), which in the solution going. The oxidation process takes place at the anode.
4. ΔE = E_{Kathodenreaktion} – E_{Anodenreaktion}
5. Ag⁺ + e^– → Ag, E^o=0.8 V, z=1
     · Reduced form Ag, c(Ag) = 1
     · Oxidated [translation unavailable] Ag⁺ , c(Ag⁺)= 0.1 mol/L
     · E₁ = 0.8 + 0.059/1 · log(0.1/1)= 0.741 V
     · Or .. E₁ = 0.8 – 0.059/1 · log(1/0.1)= 0.741 V
6. Ag⁺ + e^– → Ag, E^o=0.8 V, z=1
     · Reduced form Ag, c(Ag) = 1 mol/L
     · Oxidated [translation unavailable] Ag⁺ , c(Ag⁺)= 0.01 mol/L
     · E₂ = 0.8 + 0.059/1 · log(0.01/1)= 0.682 V
     · Or .. E₂ = 0.8 – 0.059/1 · log(1/0.01)= 0.682 V
7. It follows from the previous calculations:
     · [Ag⁺]= 0.1 mol/L → E₁=0.741 V
     · [Ag⁺]= 0.01 mol/L → E₂=0.682 V
     · ΔE = E₁ – E₂ = 0.06 V
8. Zn → Zn²⁺ + 2e^–, E^o=-0.76 V, z=2
     · reduced form Zn, c(Zn) = 1 mol/L
     · Oxidated [translation unavailable] Zn²⁺ , c(Zn²⁺)= 0.001 mol/L
     · E = E^o + 0.059/2 · log(0.001/1) = -0.849 V
October 16, 2026 at 4:42 am in reply to:
Joule, applicazioni 2 (thermo-03)
Per salire sulla Torre d'Effel occorre consumare energia che può essere fornita al corpo attraverso il cibo, quindi quante alberi di gomma si devono mangiare per compensare l'energia necessaria per salire sulla Torre d'Effel?
Nota: le dimensioni sconosciute devono essere individuate automaticamente.
#7545
Soluzione:
Diverse dimensioni devono essere rilevate o valutate:
Altezza della torre dell'Eiffel: 300 metri
Il contenuto energetico degli orsi gommosi, 100 grammi, 1472 kJ. Supponiamo che il corpo sia in grado di convertire 3/4 dell'energia. Sono disponibili circa 3/4·1472 = 1104 kJ
Massa persona (z.B. Professore di chimica) 75 kg
1. Energia necessaria per sollevare 75 kg a 300 m
  E_pot = m·g·h = 75kg · 10m/s² · 300m = 225'000 J = 225 kJ
2. Circa 1104/225 = 5 · al di sopra della torre d'Eifel rispettivamente circa 1/5 del pacco (20 grammi) per alzare la torre d'Eifel una volta
October 16, 2026 at 4:42 am
in reply to:
Redox, redux table, exercise 2 (Redox-29)
Use the redox table to assess whether the following reactions are possible.
Ag in copper sulphate
Silver nitrate in copper
You react (elementary) bromine with iron. Bromine is in excess.
[translation unavailable] in 1 M HCl
How can silver be extracted from nitrate in solid form? Only acids or bases are permitted as auxiliaries.
Gold is to be dissolved in a solution. Please specify a suitable variant.
#9507
Solution:
1. Copper sulphate: CuSO₄ thus Cu²⁺ and SO2-4
     SO2-4 Viewers, not participating
     Ag + Cu²⁺ → Ag⁺ + Cu ??
     Redux table from left -> right to interpret
     Voluntary mountain reaction not possible
2. Silver nitrate: AgNO₃ meaning Ag + and NO–3
     NO–3 Viewers not participating
     Cu + Ag⁺ → Cu²⁺ + Ag ??
     Redux table from left -> right to interpret
     Mountaineering, voluntary reaction
3. Extract from the electrode potential table:
     Fe → Fe²⁺ (-0.44)
     2·Br^– → Br₂ (+1.07)
     thus: Fe + Br₂ → 2·Br^– + Fe²⁺
     But … it goes on and on .
     Fe²⁺ → Fe³⁺ (+0.75)
     thus: Fe + Br₂ → 2·Br^– + Fe³⁺
     Load balancing: 2·Fe + 3·Br₂ → 6·Br^– + 2·Fe³⁺
4. As a reminder: 1 M HCl
     → Cl^– (viewers)
     → H₃O⁺
     Mg → Mg²⁺ (-2.37 V)
     H₂ + 2·H₂O → 2·H₃O⁺ (0.00)
     Mg + 2·H₃O⁺ → H₂ + 2·H₂O + Mg²⁺
     Mountaineering, voluntary reaction
     Magnesium dissolves to form H₂ gas
5. Silver nitrate: AgNO₃ meaning Ag + and NO–3
     Required: Ag⁺ → Ag
     In our case , Ag⁺ is the oxidizing agent !
     Reducing agents are sought (acid or base)
     However, only the base (OH^-) is available.
     4 OH^– → O₂ + 2·H₂O (0.4 V)
     Ag → Ag⁺ (0.8 V)
     Reaction occurring: 4 OH^– + Ag⁺ → O₂ + 2·H₂O + Ag
     According to the Redox table, there are also the following voluntary mountain breaks under gold:
     Au → Au³⁺ (1.42 V)
     Mn²⁺ + 12 H₂O → MnO–4 + 8·H₃O⁺ (1.51 V)
     PbSO₄ + 5 H₂O → PbO₂ + HSO–4 + 3·H₃O⁺ (1.68 V)
     2 F^– → F₂ (2.87 V)
     According to exercise a solution is to be manufactured. The reaction with F₂ would be theoretically possible, but a salt is formed. Therefore, only the variants with 1.51 V or 1.68 V are correct.
     Au + MnO–4 + 8·H₃O⁺ → Au³⁺ + Mn²⁺ + 12 H₂O … Or
     Au + PbO₂ + HSO–4 + 3·H₃O⁺ → Au³⁺ + PbSO₄ + 5 H₂O
October 16, 2026 at 4:42 am in reply to:
Calculer l’enthalpie de réaction 9 (thermo-21)
Dans un rouleau d'affichage (diamètre 6 cm, longueur 72 cm), quelques gouttes de pentane (1/4 ml) sont administrées et inflammées avec le mélange air-pentane.
a) Calculer l'enthalpie de réaction.
b) L'expérience montre que trop de pentane ne produit pas une meilleure explosion que le mélange air-pentan calculer la quantité maximale de pentane.
Dans le contexte: une expérience merveilleuse pour l'enseignement.
Remarque: des molécules de taille inconnue se produisent elles-mêmes.
#5577
Solution:
Remarque: [traduction manquante] C₅H₁₂, M(C₅H₁₂), Dichte(C₅H₁₂) = 0.63 g/ml
1. Les hypothèses:
     Péntan C₅H₁₂, M(C₅H₁₂) = 72 g/mol
     Dichte(C₅H₁₂) = 0.626 g/ml
     dHf0(C₅H12) = -147 kJ/mol
     Tous les substances gaseux concernés
     Expérience à 25℃ (293 K)
2. a) équation de réaction:
  1·C₅H₁₂ (g) + 8·O₂ (g) ⇄ 5·CO₂(g) + 6·H₂O(g)
3. ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
  ΔHoR = (5· ΔHof(CO₂,g) + 6· ΔHof(H₂O, g)) – ( ΔHof(C₅H₁₂, g) + 8· ΔHof(O₂,g))
  ΔHoR = (5·-393 + 6·-242) – (-147 + 8·0) = -3270 kJ/mol
4. l'énergie réelle libérée:
  Dichte(C₅H₁₂) = 0.626 g/ml
  Donc 1/4 ml: 0.15 grammes (=0.0021 mol)
  E = – 6.81 kJ
5. b) calcul maximum la quantité [traduction manquante]
6. Le volume du rouleau:
  V = pi · r² · h = 3.141 · 0.03² · 0.72 = 0.00204 m³
  V env. 2 litres
7. Gaz parfait 1 mol à …
     0℃ (273 K) 22.4 l
     25℃ (298 K): 24 litres
     2 litres (vol. rouleau d'affichage), soit 1/12 mol
8. L'air est constitué de 20% d'O₂, 80% de N₂.
     ainsi 1/60 mol O₂ dans le rouleau de poster
     En raison de l'équation de réaction 8· moins C₅H₁₂ que O₂, c'est-à-dire 1/480 mol C₅H₁₂
     Avec M(C₅H₁₂) = 72 g/mol, c'est suivi de 0.15 grammes C₅H₁₂
     Avec Dichte(C₅H₁₂) = 0.63 g/ml suivant: donc env. 1/4 ml C₅H₁₂
On libère donc 6.81 kJ. La quantité maximale de pentane est d'environ 1/4 ml.
October 16, 2026 at 4:42 am
in reply to:
Joule, applicazioni 1 (thermo-02)
Supponiamo di avere a disposizione 1 kJ (1000 joule).
Quanti gradi Celsius può riscaldare un litro d'acqua?
Nota: per riscaldare 1 grammo di acqua a 1°C, occorrono circa 4.18 joule.
Per quanto tempo brucia una lampada con l'indicazione di 100 watt?
Che massa può essere accelerata da 0 m/s a 1 m/s. E_kin = ½ · m · v²
Quanti battiti e per quanto tempo il cuore può battere? Supponiamo che la potenza del cuore sia di 1 watt.
In caso di successiva esercizio ('esplosione di pentano') vengono rilasciati 6.8 kJ. Quanti metri un insegnante di chimica potrebbe essere sollevato con questa energia?
#7543
Soluzione:
1. Il riscaldamento per circa 0.239 ℃
2. La data di 100 watt significa 100 joule al secondo, quindi la lampadina può funzionare per 10 secondi.
3. E_kin = ½ · m · v²
  1000 J = ½ · x · 1²
  x = 2000 kg
4. 1 watt = 1 joule al secondo. Così il cuore può battere per 1000 secondi (circa 1/4 ora).
5. E_pot = m·g·h
     Professore di chimica 68 kg (circa !)
     con g = 10 m/s²
     h = E_pot /(m·g) = 6800 J/(68 kg·10m/s²⁾ = 10 metri
October 16, 2026 at 4:42 am in reply to:
Calculer l’enthalpie de réaction 8 (thermo-20)
Parallèlement exercice à thermo-18. Il devrait mais seulement les énergie libre de Gibbs de réaction à température corporelle pour les ventilation (combustion) de l'éthanol calculé sont.
#5575
Solution:
Tableau des valeurs:
ΔHof(kJ·mol^-1) S^o(J·K^-1·mol^-1)
EtOH(l) -277 161
O₂ (g) 0 205
CO₂ (g) -393 214
H₂O (l) -286 67
1. équation de réaction:
  1·C₂H₅OH + 3·O₂ ⇄ 2·CO₂(g) + 3·H₂₀(l)
2. ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
  ΔHoR = (2· ΔHof(CO₂,g) + 3· ΔHof(H₂O, l)) – ( ΔHof(C₂H₅OH, l) + 3· ΔHof(O₂,g))
  ΔHoR = (2·-393 + 3·-286) – (-277 + 3·0) = -1367 kJ/mol (par exemple: à l'intérieur exercice thermo-107)
  ΔSoR = (2·S^o(CO₂,g) + 3·S^o(H₂O, l)) – (S^o(C₂H₅OH, l) + 3·S^o(O₂,g))
  ΔSoR = (2·214 + 3·67) – (161 + 3·205) = -147 J·mol^-1·K^-1
  ΔGoR = ΔHoR – T· ΔSoR
  ΔGoR = -1367 – (273+37)·-0.147 = -1321.43 kJ/mol
Le résultat est un peu moins à l'exergonique comment les d'autres calcul le exercice 107. Quintessence: la quantité d'alcool nécessaire serait d'environ 4% plus élevée.
October 16, 2026 at 4:42 am in reply to:
Joule, applicazioni 0 (thermo-01)
Approvato, si avrebbe 1 [traduzione mancante] a disposizione. Cosa si può fare con tutto questo?
#7541
Soluzione:
1. un cioccolato (circa 100 grammi o 0.1 kg) alzato di un metro
  E = m·g·h = 0.1kg · 10m/s² · 1m = 1 J
2. Un grammo di acqua riscaldato a 0.239 K
3. Un battito cardiaco umano consuma circa 1 J
4. Accelerare una massa di 2 kg da fermo a v=1m/s
  E =½ · m · v²
October 16, 2026 at 4:42 am in reply to:
Iron bromide (Redox-27)
After a test, a solution of zinc sulfate is contaminated with copper ions. How can the copper ions be removed efficiently? Why does a distillation and filtration not work?
#9505
Solution:
1. These are solutions (with water) and not two different solvents such as alcohol and water.
2. Filtration is only possible if a solid and a liquid are present. In this case, they are only soluble in water, individual ions. These are much too small in size compared to the pore size of a filter
3. Look at the electrode potential table:
     Zn … Zn²⁺ (-0.76 V)
     Cu … Cu²⁺ (+0.35 V)
     So if a Zn plate is in …
October 16, 2026 at 4:42 am in reply to:
Ossidoriduzione, elettrolisi, 8 (Redox-58)
Da una soluzione di sulfato di rame si deve eliminare 800 mg di rame a I=180mA alla catode. Calcolare a) la durata del tempo di elettrolisi e b) determinare il volume di ossigeno che si produce contemporaneamente in condizioni normali (298 K, 101300 N/m²).
La equazione generale dei gas è p·V=n·R·T, con R = 8.314 J/molK.
Nota : identici esercizio anche al momento della tematica sali.
#7539
Soluzione:
1. · Coinvolti reazioni:
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 = circa 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 = circa 154 ml
14. · Unità: [mol·J/(mol·K)·K / N/m²]
October 16, 2026 at 4:42 am
in reply to:
Calculer l’enthalpie de réaction 7 (thermo-19)
Adoptée, on décide se pour les pas en bonne santé [traduction manquante], que les complète quotidiennement réservation de l'énergie par le biais de boisson de l'alcool (40% boissons alcoolisées) réalisé est. Les hypothèses et les spécifications suivantes s'appliquent:
L'alcool de boisson serait C₂H₅OH (éthanol), la densité de l'alcool pur est de 0.9 g/ml
Supposons que l'alcool soit inhalé dans le corps, c'est-à-dire mis en œuvre avec O₂.
Remarque 1: Calculer d'abord l'enthalpie de réaction, y compris l'état d'agrégation
Remarque 2: calculer à partir de ΔHoR et de la quantité d'éthanol par litre d'alcool la quantité nécessaire pour réaliser l'approvisionnement en énergie
Remarque 3: La consommation énergétique quotidienne est de 9200 kJ
Remarque 4: pour les professionnels: les base de calcul est pas entièrement, en fait il faudrait les énergie libre de Gibbs de réaction à température corporelle calculé sont. Voir détails les suivants exercice thermo-108. Mais vous pouvez voir que le résultat ne changerait que d'environ 10%
#5573
Solution:
1. équation de réaction:
  1·C₂H₅OH + 3·O₂ ⇄ 2·CO₂(g) + 3·H₂O(l)
2. ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
  ΔHoR = (2· ΔHof(CO₂(g)) + 3 ΔHof(H₂O(l))) – ( ΔHof(C₂H₅OH) + 3· ΔHof(O₂))
  ΔHoR = (2·-393 + 3·-286) – (-278 + 3·0) = -1366 kJ/mol
  Donc le savoir on [traduction manquante], que par Mol l'éthanol 1366 kJ libéré sont
3. La prochaine question à résoudre: combien de grammes d'éthanol y a-t-il dans l'alcool ?
  1 litre d'alcool, 40%, soit 400 ml d'éthanol pur. Avec une densité de 0.9 g/ml, 360 grammes d'éthanol sont ainsi produits.
4. M(C₂H₅OH) = 46 g/mol, donc
  n= m/M = 360 / 46 = 7.8 mol
5. Par bouteille, les alcools contiennent donc 7.8 mol à -1366 kJ, soit -10690 kJ.
6. Pour un besoin quotidien de 9200 kJ besoin il donc env. 860 ml
October 16, 2026 at 4:42 am in reply to:
Redox, redux table, Herleitung-4 (Redox-26)
Alternative derivation of the previous equation.
a) H₂ + 2·H₂O → 2·H₃O⁺ (pH=7, -0.42 V)
b) 4·OH^– → O₂ + 2·H₂O (pH=7, 0.83V)
The reaction would be mountaintop or lower left to upper right. the whole reaction with the indication that the reaction ·2 should be made.
The second question … [translation unavailable] ·2
#9503
Solution:
1. a) H₂ + 2·H₂O → 2·H₃O⁺ ( + 2·e^– )
2. b) 4·OH^– → O₂ + 2·H₂O ( + 4·e^– )
3. (b) + 2·a, rock reaction, electron balance
     4·OH^– + 4·H₃O⁺ → 2·H₂ + 4·H₂O + O₂ + 2·H₂O
     4·(OH^– + H₃O⁺) → 2·H₂ + O₂ + 6·H₂O
     4·( 2·H₂O ) → 2·H₂ + O₂ + 6·H₂O // – 6·H₂O
     2·H₂O → 2·H₂ + O₂
October 16, 2026 at 4:42 am in reply to:
Redox, redux table, Herleitung-3 (Redox-25)
In the cathode space (pH=0) the reaction equation is
4·H₂O + 4·e^– → 2·H₂ + 4·OH^–
and in the anode space (pH=14)
6·H₂O → O₂ + 4·H₃O⁺ + 4·e^–
Simplify as much as possible !
#9501
Solution:
1. a) 4·H₂O + 4·e^– → 2·H₂ + 4·OH^–
2. b) 6·H₂O → O₂ + 4·H₃O⁺ + 4·e^–
3. a) + b)
  4·H₂O + 6·H₂O + 4·e^– → O₂ + 4·H₃O⁺ + 4·e^– +2·H₂ + 4·OH^–
  10·H₂O → O₂ + 2·H₂ + 4·H₃O⁺ + 4·OH^–
4. with 2·H₂O → O₂ + 2·H₂ // subtract both sides
  8·H₂O → 4·H₃O⁺ + 4·OH^–
5. with H₃O⁺ + OH^– → 2·H₂O
  8·H₂O → 6·H₂O + 2·H₂ + O₂
6. 2·H₂O → 2·H₂ + O₂
October 16, 2026 at 4:42 am in reply to:
Ossidoriduzione, elettrolisi, 7 (Redox-57)
Calcola lo spessore del strato di rame che un flusso di 5.8 A emette uniformemente in 3 ore su una sfera con un raggio di 8.3 cm.
Nota 1: la superficie di una sfera è 4·pi·r²
Nota 2: La copertura è stata effettuata con una soluzione di sulfato di rame
Nota 3: Dichte(Cu) = 8.9 g/cm³
Nota : identici esercizio anche al momento della tematica sali.
#7537
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
10. → d = m/(Dichte·4·π·r²)
11. d = 20.63/(8.9·4·π·8.3²) = 2.67·10^-3cm = 0.026 mm
October 16, 2026 at 4:42 am in reply to:
Calculer l’enthalpie de réaction 6 (thermo-18)
Calculer le ΔHof du monoxyde de carbone. Considérez la réaction suivante:
CH₄(g) + H₂O(l) → CO(g) + 3 H₂(g). Les (d'expérimentation) ΔHoR pour cela à hauteur 249.5 kJ/mol.
#5571
Solution:
ΔHoR = [ ΔHof(CO(g)) + 3· ΔHof(H₂(g))] – [ ΔHof(CH₄) + ΔHof(H₂O(l))] ΔHoR = 249.5 kJ/mol = [x + 3·0]-[-74.85 + -285.83]
x = …
October 16, 2026 at 4:42 am in reply to:
Calculer l’enthalpie de réaction 5 (thermo-17)
On a besoin d'un gaz à combustion pour transpirer et couper. La plupart du temps prend on à cet effet éthylène (C₂H₂). [traduction manquante] prend on pas le moins cher le méthane (CH₄)?
Remarque: formulez la réaction de combustion pour les deux gaz et déterminez l'enthalpie de réaction par molécule C₂H₂ et CH₄ !
ΔHof(CH₄) = -75 kJ/mol
ΔHof(C₂H₂) = 227 kJ/mol
ΔHof(H₂O) = -285 kJ/mol
ΔHof(CO₂) = -393 kJ/mol
#5569
Solution:
1. Brûlure d'éthine:
  1·C₂H₂(g) + 5/2·O₂(g) → 1·H₂O(g) + 2·CO₂(g)
  ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
  ΔHoR = (1·-285+2·-393)-(1·227+5/2·0) = -1298 kJ/mol
2. Brûlure de méthane:
  1·CH₄(g) + 2·O₂(g) → 2·H₂O(g) + 1·CO₂(g)
  ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
  ΔHoR = (2·-285+1·-393)-(1·-75+2·0) = -880 kJ/mol
Le calcul montre qu'environ 1.5 fois plus d'énergie par vol d'un mol d'éthine est libéré par rapport au méthane.
October 16, 2026 at 4:42 am in reply to:
Redox, redux table, Herleitung-2 (Redox-24)
In the cathode space (pH=0) the reaction equation is
2·H₃O⁺ + 2·e^– → H₂ + 2·H₂O
Forms such that they appear on the right-hand side 4·OH^–.
#9499
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^–
October 16, 2026 at 4:42 am in reply to:
Ossidoriduzione, elettrolisi, 6 (Redox-56)
Una scaffo con una superficie di 120 cm² dovrebbe attraverso elettrolisi una soluzione di nitrato d'argento un 0.2 mm spessore trasporto d'argento ricevuti. Calcolare il tempo necessario se la potenza elettrica è di 10 A. La densità dell'argento è 10.5 g/cm³.
Nota : identici esercizio anche al momento della tematica sali.
#7535
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
October 16, 2026 at 4:42 am in reply to:
Ossidoriduzione, elettrolisi, 5 (Redox-55)
Per quanto tempo deve fluire un flusso di 10 A (circa il solito 'acquisto') fino a quando elettroni di massa totale di 1 kg sono passati attraverso il conducente?
Nota m(e^–) = 9.1·10^-31 kg, carico 1e^– = 1.602·10^-19 C
Nota : identici esercizio anche al momento della tematica sali.
#7533
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
Author
Posts

	à base	acide conique
H₂O	OH^–	H₃O⁺
HS^–	S^2-	H₂S
F^–	x	HF
H₂PO–4	HPO2-4	H₃PO₄
PO-34	x	HPO2-4
H₃PO₄	H₂PO–4	x
NH₃	NH–2	NH+4
NO–2	x	HNO₂

substance	M(g/mol)	m(g)	n(mol)
H₂SO₄	98	276	2.816
Pb	207.2	291.77	1.408
PbO₂	239.2	336.79	1.408

c(Cu²⁺)	c(Zn²⁺)	ΔE
1	1.0	1.11
0.9	1.1	1.107
0.8	1.2	1.105
0.6	1.4	1.099
0.4	1.6	1.092
0.2	1.8	1.082
0.001	1.999	1.012

Viewing 49 posts - 197 through 245 (of 1,381 total)

← 1 2 3 4 5 6 … 27 28 29 →

	ΔHof(kJ·mol^-1)	S^o(J·K^-1·mol^-1)
O₂ (g)	0	205.03
N₂ (g)	0	191.61
NOO (g)	90.37	210.62

	ΔHof(kJ·mol^-1)	S^o(J·K^-1·mol^-1)
H₂ (g)	0	131
O₂ (g)	0	205
H₂O (g)	-242	189
H₂O₂ (g)	-136	233

	ΔHof(kJ·mol^-1)	S^o(J·K^-1·mol^-1)
EtOH(l)	-277	161
O₂ (g)	0	205
CO₂ (g)	-393	214
H₂O (l)	-286	67