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October 16, 2026 at 4:39 am in reply to:

Molécules dessiner, charge formelle

(Atom-48)

Charge formelle: déterminer le charge formelle dans les molécules.

#5193

Solution:

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Mitschrift:
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October 16, 2026 at 4:39 am in reply to:

Concentration, density 2 (st-96)

Calculate the following densities:

Calculate the density of the following gases: O₂, N₂ and CO₂ under normal conditions
Calculate the density of air
Explain why a fire can be extinguished with a CO₂ extinguisher.

#9123

Solution:

Note: Normal conditions for gases are: one mol takes 22.4 L
Thus: n = 1, V = 22.4 L
From density=m/V and n=m/M following:: D=n·M/V

Density = m/V = n·N/V
   Dichte(CO₂)= 1mol·M(CO₂)/22.4 L = 44/22.4= 1.96 g/L
   Dichte(O₂) = 32/22.4 = 1.43 g/L
   Dichte(N₂) = 28/22.4 = 1.25 g/L
Air composed of 80% N₂ and 20% O₂
   Component N₂: 80% of a mol: 0.8 mol
   Component O₂: 20% of a mol: 0.2 mol
   Dichte(Luft) = (0.8·M(N₂)+0.2·M(O₂))/22.4 = (0.8·28+0.2·32)/22.4 = 1.29 g/L
The density of CO₂ is 1.96 g/L, of which air is 1.29 g/L. Thus CO₂ is 'heavier' than air and thus suppresses the 'lighter' O₂ (thickness 1.43 g/L) which stifles the flame as no oxygen can penetrate the candle flame.

October 16, 2026 at 4:39 am in reply to:

Legge di Coulomb, applicazione 2 (Atom-21)

La forza tra un elettrone e un protone attira o respinge?

#7159

Soluzione:

Le due particelle hanno visto la stessa carica dal punto di vista della dimensione, ma queste dimensioni sono opposte dal punto di vista del segnale. I protoni sono carichi positivi, gli elettroni negativi. Quindi queste due particelle si attraggono.

October 16, 2026 at 4:39 am in reply to:

Coulombkraft – Gravitationskraft (Salze-51)

Jemand behauptet, dass die Anziehungskräfte in einem Salz vor allem durch die Gravitationskräfte (F_G) der beteiligten Ionen bestimmt werden. Um wie viel stärker ist die Anziehung aufgrund der Coulombkraft (F_c) im Vergleich zur Gravitationskraft (F_G)?
Nimm für die Berechnung die Anziehungskräfte zwischen einem Natriumion und einem Chloridion. Der Abstand der beiden Ionen betrage 2.82·10^-10 m.

#3423

Lösung:

F_c = k·Q₁·Q₂/r²
   Q₁ = Q₂ = Elementarladung = 1.60222·10^-19 C
   F_c = (8.9875·10⁹)·(1.6022·10^-19)² / r²
   F_c = 2.31·10^-28 / r² [N]
F_G = G·m₁·m₂/r²
   · Einzelne Atome nehmen … 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]
Faktor x = F_c/F_G
x = (2.31·10^-28 / r² ) / (1.5·10^-61 / r² ) = 1.5·10³³

Der Abstand der Ionen kürzt sich in der Gleichung heraus. Als Resultat kann gesagt werden, dass die Coulombkraft eindeutig stärker ist als die Gravitationskraft.

October 16, 2026 at 4:39 am in reply to:

Concentration, density 1 (st-95)

One solution has a mass of 120 grams. Its volume is 110 ml. Calculate its density.

#9121

Solution:

Density = m/V
d = 120 g / 110 ml = 1.09 g/ml

October 16, 2026 at 4:39 am in reply to:

Legge di Coulomb, applicazione 1 (Atom-20)

Per due particelle con cariche opposte, la distanza è raddoppiata.

#7157

Soluzione:

La legge di Coulomb afferma che la forza tra due carichi è direttamente proporzionale al prodotto delle cariche e viceversa proporzionale al quadrato della distanza tra loro. Se si raddoppia la distanza tra le cariche, la forza viene ridotta di un fattore di 1/4, oppure: se si raddoppia la distanza, la nuova forza è un quarto della forza originale.

L'intera situazione è più matematica: F_a è la forza che agisce prima del cambiamento, F_b è la forza dopo il cambiamento.

F_a =k·Q₁·Q₂ / x²
F_b =k·Q₁·Q₂ / (2x)²
   = k·Q₁·Q₂ / (4x²)
   = k·Q₁·Q₂ / x² · 1/4
   = F_a · 1/4

October 16, 2026 at 4:39 am in reply to:

Dessin des molécules, angles de liaison

(Atom-47)

Angle de liaison: comment devraient les molécules dessiné sont, [traduction manquante] les angle de liaison correctement sont. On suppose que le dessin des molécules est maîtrisé.

#5191

Solution:

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October 16, 2026 at 4:39 am in reply to:

Concentration, exercise 4 (st-94)

Different calculations around the concentration.

To 150 ml of a solution with c = 0.3 mol/l, 200 ml of water is given.
To 300 ml a solution with c = 0.3 mol/l are 200 ml a (equal to) solution with c= 0.5 mol/L given. What concentration does the solution have now?
Give 1.5 L of a boiling salt solution with c = 0.15 mol/L.
Well, you know, should but 2.0 L a cooking salt solution with c = 0.3 mol/L manufacture are. How many grams of salt and water must be added to the given saline solution.

#9119

Solution:

c = n/V or n = c·V
   n(Lösung) = c·V = 0.3 mol/l·0.15 l = 0.045 mol
   new volume: 150 ml + 200 ml = 350 ml or 0.35 L
   c = n/V = 0.045 mol / 0.35 L = 0.129 mol/L
c = n/V or n = c·V
   n₁(Lösung) = c₁·V₁ = 0.3 mol/l·0.3 l = 0.09 mol
   n₂(Lösung) = c₂·V₂ = 0.5 mol/l·0.2 l = 0.1 mol
   new n: n₁ + n₂ = 0.19 mol
   new volume: 300 ml + 200 ml = 500 ml or 0.55 L
   c = n/V = 0.19 mol / 0.5 L = 0.38 mol/L
Additional volume: 2.0 L – 1.5 L = 0.5 L
   n₁(1.5 L Lösung) = c₁·V₁ = 0.15 mol/L·1.5L = 0.225 mol
   n₂(2.0 L Lösung) = c₂·V₂ = 0.3 mol/L·2L = 0.6 mol
   with n=m/M or m(NaCl) =n·M(NaCl)
   M(NaCl) = 58.5 g/mol
   m₁(NaCl) =0.225 mol · 58.5 g/mol = 13.16 g
   m₂(NaCl) =0.6 mol · 58.5 g/mol = 35.1 g
   Additional amount of NaCl = 35.1-13.16 = approximately 22 grams

October 16, 2026 at 4:39 am in reply to:

Isotopi: frequenza percentuale 6, struttura atomica (Atom-19)

L'argento si trova come miscela di due isotopi, ¹⁰⁷Ag, massa atomica 106.906 u e ¹⁰⁹Ag con massa atomica 108.905 u. Qual è la percentuale di partecipazione di ogni isotopo? Calcolo solo fino a a al punto, fino a le equazione solo ancora una sconosciuti maggiore ha.

#7155

Soluzione:

Isotopo I: ¹⁰⁷Ag, 106.906 u, x %
Isotopo II: ¹⁰⁹Ag, 108.905 u, y %
x % + y % = 100 %
y % = 100 % – x %
PSE: m(₄₇Ag) = 107.8662 u
x·106.906 u + (1-x)·108.905 u = 107.8662 u
non richiesto, ma come esercizio, dopo risolvere x dà un valore di 51.966% per il isotopo I
E 48.034% per il isotopo II

October 16, 2026 at 4:39 am in reply to:

Les molécules sont dessinées 4

(Atom-46)

Dessin de molécules II, de liaison simple et multiples: Dessin d'autres molécules, même avec des liaisons doubles et triples.

#5189

Solution:

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October 16, 2026 at 4:39 am in reply to:

Lösung mit gegebener Konzentration herstellen (Salze-50)

Stelle 0.4 Liter eine 0.2 mol/l NaCl-Lösung her im Labor mit einer Waage und Messkolben her.

#3421

Lösung:

c=n/V=m/M/v –> m =c·V·M = 0.2 mol/l·0.4 l · 58.5 g/mol = 4.68 g
Herstellung: 4.68 g NaCl werden in ein Becherglas gegeben. Nun wird mit Wasser bis an die Marke 0.4 Liter aufgefüllt.
Die Antwort: es werden nun 0.4 Liter hinzugefügt ist nicht vollständig korrekt, weil dies (minimal) nicht die gleiche Menge ist wie wenn man bis auf die Marke 0.4 Liter auffüllt: das NaCl beansprucht ein kleines Volumen.

October 16, 2026 at 4:39 am in reply to:

Frequenza percentuale 5, struttura atomica (Atom-18)

Un elemento è costituito da una miscela di tre isotopi, i cui due isotopi più leggeri hanno le stesse frequenze. L'masse molari dei tre isotopi è anche noto.
Calcola di cui: le generale formula il medie massa molare del elementi.

#7153

Soluzione:

Frequenza isotopo I: x, in cui: x una numero tra: 0 (0%) e 1 (100%) sarebbe
Frequenza isotopo II: x
Quindi l'isotopo III ha la frequenza (1-2x)
A attesa massa molare: x·M(Isotop I) + x·M(Isotop II) + (1-2x)·M(Isotop III)

October 16, 2026 at 4:39 am in reply to:

Concentration, exercise 10 (st-93)

Calculate the concentration of household sugar (C₁₂H₂₂O₁₁), in a cup coffee (150 ml), where: [translation unavailable] a cabbage sugar (m = 7.7 g) dissolved is.
Addition: How many molecules of caffeine (C₈H₁₀N₄O₂) are present in the body after a cup of coffee, when in a Cup coffee 80 mg of which: dissolved were?

#9117

Solution:

c = n/V = m/M/V
M(C₁₂H₂₂O₁₁) = 342 g/mol
c = 7.7g/342g/mol / 0.15 L = 0.15 mol/L
Additional task:
n = m/M
   M(C₈H₁₀N₄O₂) = 194 g/mol
   · n = 0.08/194 = 0.000412 mol = 4.12·10^-4 mol
   · 1 mol = 6.022·10²³
   · 4.12·10^-4 mol → 2.48·10²⁰

October 16, 2026 at 4:39 am in reply to:

Mehratomige Anionen zeichnen (Salze-40)

Zeichne folgende mehratomigen Anionen:

Sulfat
Nitrat
Carbonat
Hydrogensulfat
Phosphat

Hinweis: Nach Möglichkeit machen die Sauerstoffatome keine Bindung mit sich selbst.

#3419

Lösung:

Es handelt sich dabei um mehratomige Anionen:

Sulfat: SO2-4
Nitrat: NO–3
Carbonat: CO2-3
Hydrogensulfat: HSO–4
Phosphat: PO3-4

Die gezeichneten Lewisstrukturen findet man im Video. noch nichts gemacht

October 16, 2026 at 4:39 am in reply to:

Molécules dessiner, règle de l’octet

(Atom-45)

Règle de l’octet (y compris à caractère obligatoire – pas à caractère obligatoire couples d'électrons): comment peut la raison une dessinées la structure tout simplement et rapidement décidés sont, ou le molécule en principe correctement est?

#5187

Solution:

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October 16, 2026 at 4:39 am in reply to:

Dessin des molécules 2

(Atom-44)

Liaison simple: dessinez les molécules H₂O, CH₄, H₂O₂, NH₃, C₂H₆O (deux différentes les variantes)

#5185

Solution:

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Erklärung in Form eines Videos, Dauer 15:00

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October 16, 2026 at 4:39 am in reply to:

Concentration, exercise 12 (st-92)

1 grams [translation unavailable] should on the oceans and seas distributed are. How many gold atoms are there per liter?

#9115

Solution:

Idea: about 60% of the Earth's surface is covered by water, Earth radius about 6000 km, average water depth about 3 km
All these assumptions provide a volume of 0.814·10⁹ km³
Wikipedia has almost the same volume:
V = 1.338·10⁹ km³ = 1.338·10²¹ litres
The following is calculated using the Wikipedia value
1 g Au → n(Au)=1g/197g/mol= 1/197 mol
1/197 mol ≙ 3.057·10²¹ Au-atoms
3.057·10²¹ / 1.338·10²¹ = 2.28 atoms per litre

October 16, 2026 at 4:39 am in reply to:

Magnesiumtetroxid (Salze-32)

Jemand behauptet, dass MgO₄ ein Wundermittel gegen Dummheit wäre. Lohnt sich ein Kauf?

#3417

Lösung:

Es gibt keine bekannte Verbindung mit der chemischen Formel MgO₄. Magnesium bildet in der Regel Verbindungen mit einer Oxidationszahl von +2, was bedeutet, dass Magnesiumionen (Mg²⁺⁾ vorliegen. Daher wäre Verbindung Magnesiumoxid (MgO) zu erwarten.

Zudem: Beim MgO₄ müsste das Magnesium eine Ladung von 8+ haben, da die Sauerstoffionen ja ein Ladung von jeweils -2 haben. Ein Blick auf die üblichen Ladungen zeigt, dass irgendwo bei einer Ladung von ca. 5+ das Atom derart positiv geladen ist, dass es Elektronen aufnimmt, egal woher und somit seine (positive) Ladung verkleinert.

Quintessenz: Ein Kauf lohnt sich nicht.

October 16, 2026 at 4:39 am in reply to:

Frequenza percentuale 4, struttura atomica (Atom-17)

Il bromo si presenta in natura come miscela dei due isotopi stabili ⁷⁹Br e ⁸¹Br. Frequenze: ⁷⁹Br a 50.65%, m = 78.918338 u. Calcolare dalla media massa molare del freno ora la frequenza e la massa del ⁸¹Br in u e g/mol.

#7151

Soluzione:

Frequenza del il secondo isotopi: 100 – 50.65% = 49.35%
Quindi: 0.5065·78.918338 u + 0.4935· x u = 79.904 u
x = 80.9156 u, rispettivamente 80.9156 g/mol

October 16, 2026 at 4:39 am in reply to:

Frequenza percentuale 3, struttura atomica (Atom-16)

L'elemento cloro è composto da due isotopi: ³⁵Cl e ³⁷Cl. m(³⁵Cl) = 34.969 u, frequenza = 75.77%, m(³⁷Cl) = 36.966 u.
Calcolare la massa media di un cloratomo e interpretare il risultato.

#7149

Soluzione:

m(Cl) = 0.7577·34.969 u + (1-0.7577)·36.966 u = 35.4528731 u
Le media massa di un cloratomi è 35.4528 u oppure Un Mol avrebbe la massa di 35.4528 grammi. Ricorda: anche se cerchiamo così bene nella natura, non troveremo mai un atomo del genere. O i cloratomi hanno la massa di 34.969 u o la massa di 36.966 u.

October 16, 2026 at 4:39 am in reply to:

Concentration, exercise 1233 (st-90)

Different calculations around the concentration.

5 grams of salt is added to a liter of water. What concentration does the solution have?
18 g C₆H₁₂O₆ is dissolved in 120 ml of water. What concentration does the solution have?
10 ml of a 0.02 mol/l vitamin C solution were produced.
Notes: M(Vitamin C) = 176.13 g/mol
100 ml of magnesium bromide solution (MgBr₂) with a concentration of c = 0.02 mol/L is to be produced.
How can this solution be made in a chemical laboratory?
One liter of a salt solution weighs 1.34 kg. What concentration does the solution have?

For the record: c = n/V = m/M/V

#9113

Solution:

The volume must be expressed in litres.

c = m(NaCl)/M(NaCl)/V = 5g/58.5g/mol/1.0 L = 0.085 mol/l
c = m(C₆H₁₂O₆)/M(C₆H₁₂O₆)/V = 18g/180g/mol/0.12L = 0.833 mol/l
c = n/V = m/M/V → m = c·V·M
m = 0.02 mol/l · 0.01 l · 176.13 g/mol = 0.035 g
c = n/V = m/M/V → m = c·V·M
   M(MgBr₂) = 184.3 g/mol
   m = 0.02 mol/L · 0.1 L · 184.3 g/mol = 0.3686 g
   Therefore, 0.3686 grams of MgBr₂ must be weighed and so much water added that the final volume is 100 ml.
Assumption: 1 litre of water is 1.0 kg, so the rest is salt: 340 grams
n=m(NaCl)/M(NaCl) = 340g/58.5 g/mol = 5.81 mol
c = n/V = 5.81 mol / 1.0 L = 5.81 mol/L

October 16, 2026 at 4:39 am in reply to:

Typische Salze (Salze-31)

Nenne zehn typische Salze.

#3415

Lösung:

Die Antworten können unterschiedlich sein. Es gibt tausende von Salzen. Antwort von chatgpt, inklusive dessen Geschwätzigkeit:
Salze sind Verbindungen, die aus positiv geladenen Metallionen und negativ geladenen Nichtmetallionen (Anionen) bestehen. Hier sind zehn Beispiele für typische Salze:

Natriumchlorid (NaCl): Gewöhnliches Kochsalz, das in der Küche verwendet wird.
Kaliumnitrat (KNO₃): Ein Salz, das in der Lebensmittelindustrie und als Düngemittel verwendet wird.
Calciumcarbonat (CaCO₃): Dieses Salz kommt in Kalkstein, Kreide und Marmor vor.
Eisen(II)-sulfat (FeSO₄): Ein Salz, das oft als Düngemittel und zur Wasserbehandlung verwendet wird.
Natriumcarbonat (Na₂CO₃): Auch bekannt als Soda, wird in der Glasproduktion und als Reinigungsmittel verwendet.
Natriumsulfat (Na₂SO₄): Ein Salz, das als Trocknungsmittel und in der Seifenherstellung verwendet wird.
Aluminiumoxid (Al₂O₃): Dieses Salz kommt in der Natur als Bauxit vor und wird zur Aluminiumgewinnung verwendet.
Ammoniumchlorid (NH₄Cl): Ein Salz, das in einigen Säure-Basen-Reaktionen und in der Medizin verwendet wird.
Natriumhydrogencarbonat (NaHCO₃): Auch bekannt als Natron, wird in Backpulver verwendet.
Kalziumchlorid (CaCl₂): Ein Salz, das als Festigungsmittel in Lebensmitteln und als Streusalz im Winter verwendet wird.

Diese Beispiele zeigen die Vielfalt der Salze, die in verschiedenen Bereichen wie Lebensmittelindustrie, Landwirtschaft, Chemie und Medizin Anwendung finden.
Kommentar: Die Namen inklusive der Salzformeln sind tadellos. Quintessenz: Das typische (aktuell) Erfinden von Fakten seitens chatgpt konnte nicht beobachtet werden. Ein Check mit anderen Quellen aber gut und wichtig.

October 16, 2026 at 4:39 am in reply to:

Des molécules dessinent 1

(Atom-43)

notation de Lewis: Comment les éléments sont-ils représentés dans l'orthographe de Lewis?

#5183

Solution:

Aucun texte, aucune séquence vidéo

Videoantwort:
Erklärung in Form eines Videos, Dauer 10:12

Direkter Link zum Fullscreen-Modus hier

Mitschrift:
Transcription de la vidéo au format PDF [hier]

October 16, 2026 at 4:39 am in reply to:

Various calculations, ideal gas law (st-84)

Different calculations

At a pressure of 75 kPa, a gas sample takes up a volume of 360 ml. What volume does the sample take up at the same temperature under a pressure of 100 kPa ?
The volumes a gas sample is 462 ml at 35 ℃ and 115 kPa. What 's the volume under normal conditions ?
The quantity of litres of CO (g) measured under normal conditions is required to produce 1.0 kg of rust (Fe₂O₃) by the following masses:
Fe₂O₃ + CO → Fe + CO₂
A Mol of a gas claims at normal conditions 22.4 litres. Calculate the side length of the Mol spindle.
How many pounds of air are in a classroom?
Assumptions: 180 m³, 20 ℃ 100'000 Pa, air 80% N₂, 20% O₂

#9111

Solution:

Note that the gas equation can be formulated as follows:
p·V = n·R·T → p·V/T = n·R
Note that after this reshaping, the term n·R is on one side. the value, no matter how large p, V and T are, is always the same and the following practical equation results:
p₁·V₁/T₁ = p₂·V₂/T₂
The values 1 and 2 shall indicate that this is situation 1 and 2.

Es of which: p₁·V₁/T₁ = p₂·V₂/T₂ with …
   p₁ = 75 kPa, V₁ = 360 ml, p₂ = 100 kPa, T₂ = T₁
   V₂ = p₁·V₁·T₁/(p₂·T₁) = p₁ ·V₁/p₂
   V₂ = 75 kPa · 360 ml / 100 kPa = 270 ml
Es of which: p₁·V₁/T₁ = p₂·V₂/T₂ with …
   p₁ = 115 kPa = 115'000 Pa, V₁ = 462 ml, T₁ = 35 ℃ = 308 K
   p₂ = 103'000 Pa, T₂ = 0 ℃ = 273 K
   V₂ = p₁·V₁·T₂/(p₂·T₁)
   V₂ = 115'000 Pa · 462 ml · 273 K / (308 K · 101'000 Pa) = 466 ml
Balancing the reaction equation:
   1 · Fe₂O₃ + 3 ·(g) CO → 2 · Fe + 3 · CO₂
   n(Fe₂O₃) = m/M = 1000/159.6 = 6.27 mol
   n(CO) = 3 · 6.27 = 18.8 mol
   1 mol (under standard conditions) is equivalent to 22.4 L
   18.8 mol: 421 litres
x³ = 22.4 L = 22.4 dm³ = 0.0224 m³
x = ∛(0.0224m³) = 0.28 m
Es of which: p₁·V₁/T₁ = p₂·V₂/T₂ with …
   p₁ = 101'300 Pa, V₁ = 22.4 L, T₁ = 0 ℃ = 273 K
   p₂ = 100'000 Pa, T₂ = 20 ℃ = 293 K
   V₂ = p₁·V₁·T₂/(p₂·T₁)
   V₂ = 101300·22.4·293/(273·100'000) = 24.35 L
   Dichte(Luft) = (0.8·28+0.2·32) / 24.35 = 1.18 g/L
   1.18 g/L = 1.18 kg/m³
   180 m³ → 212.4 kg

October 16, 2026 at 4:39 am in reply to:

Frequenza percentuale 2, struttura atomica (Atom-15)

Più naturale carbonio esiste a 98.94% da ¹²C, il restante esistono da ¹³C (m=13.003355 u).

Sulla base di questi valori calcoli la massa media di carbonio attesa.
Cosa si può dedurre dal valore effettivo di 12.0107 (valore PSE)?
Calcola le a attesa media massa di carbonio con il idea, che le masse di m(¹²C)=12u e m(¹³C)=13u quantità.

#7147

Soluzione:

0.9894·12 u + 0.0106·13.003355 u = 12.010635 u
Il valore teorico ('PSE') è di 12.0107 u, rispettivamente di 12.0107 g/mol.
0.9894·12 u+ 0.0106·13 u = 12.0106 u

October 16, 2026 at 4:39 am in reply to:

Le nombre possible de liaisons d’un atome d’hydrogène (Atom-40)

Combien de liaisons un atome d'hydrogène peut-il faire ?

#5181

Solution:

Un atome d'hydrogène a un électron de valence et peut donc exactement une liaison entraînement. Pour professionnels: liaisons multicentriques sont pas traitées. 🙂

October 16, 2026 at 4:39 am in reply to:

Benenne folgende Ionen und Salze (Salze-30)

Gegeben seien folgende Ionen sowie Salze. Benenne diese mit ihrem Namen. Beachte, dass bei den Übergangsmetallen die Ladung mit römischen Ziffern angegeben werden muss.

NaCl:
F^– :
SO2-4 :
O^2- :
NaNO₃ :
Al₂O₃:
FeO :
Fe₂O₃ :
NaS :
C^4- :
CaO₂:
NaHCO₃^–:
Na₂SO₄:
NH₄⁺:
PbS:
Al₄C₃:
(NH₄)(NO₃):
AlBr₃ :
Fe(NO₃)₃ :
(NH₄)₂S :
Fe₃(PO₄)₂ :
Fe(HPO₄) :
Fe(H₂PO₄)₃ :
Fe₂(SO₄)₃ :
Fe(HSO4)₂ :

#3413

Lösung:

NaCl: Natriumchlorid
F^– : Fluorid
SO2-4 : Sulfat
O^2- : Oxid
NaNO₃ : Natriumnitrat
Al₂O₃: Aluminiumoxid
FeO: Eisen(II)oxid
Fe₂O₃: Eisen(III)oxid
NaS: Natriumsulfid
C^4- : Carbid
CaO₂: Calciumoxid
NaHCO₃^–: Natriumhydrogencarbonat
Na₂SO₄: Natriumsulfat
NH₄⁺: Ammonium
PbS: Blei(II)-sulfid
Al₄C₃: Aluminiumcarbid
(NH₄)(NO₃): Ammoniumnitrat
AlBr₃ : Aluminiumbromid
Fe(NO₃)₃ : Eisen-III-nitrat
(NH₄)₂S : Ammoniumsulfid
Fe₃(PO₄)₂ : Eisen-II-phosphat
Fe(HPO₄) : Eisen-II-hydrogenphosphat
Fe(H₂PO₄)₃ : Eisen-III-dihydrogenphosphat
Fe₂(SO₄)₃ : Eisen-III-sulfat
Fe(HSO4)₂ : Eisen-II-hydrogensulfat

October 16, 2026 at 4:39 am in reply to:

Frequenza percentuale 1, struttura atomica (Atom-14)

Qual è la percentuale di ²⁸Si e ²⁹Si, se la percentuale di massa di ³⁰Si è 3.1 %?
Nota: m(²⁸Si) = 27.9769 u, m(²⁹Si) = 28.9765 u, m(³⁰Si) = 29.9738 u

#7145

Soluzione:

Il sistema periodico fornisce la massa media del silicio: 28.09 g/mol, rispettivamente. la massa per ipotetico atomo di silicio.
x = frequenza in % di ²⁸Si, y = frequenza in % di ²⁹Si
Equazione 1: x + y + 3.1 = 100
Equazione 2: (x·27.9769u + y·28.9765u + 3.1·29.9738u) / 100 = 28.09u
determinare x e y con le equazioni 1 e 2. x = 92.2 %; y = 4.7%

October 16, 2026 at 4:39 am in reply to:

Configuration électronique (Atom-35)

Comment est les complète configuration électronique suivants éléments.

C
Nitrogen
Cu
Na⁺
O^2-
Ca²⁺
Un élément de 31 électrons

#5179

Solution:

atome	nombre p,e^–	configuration électronique
C	6p, 6e^–	1s² 2s² 2p²
N	7p, 7e^–	1s²2s²2p³
Cu	29p, 29e^–	1s²2s²2p⁶3s²3p⁶4s²3d⁹
Na⁺	11p, 10e^–	1s²2s²2p⁶
O^2-	8p, 10e^–	1s²2s²2p⁶
Ca²⁺	20, 18e^–	1s²2s²2p⁶3s²3p⁶
	31e^–	1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p¹

Remarque: un coup d'œil à un PSE peut indiquer une configuration électronique différente. Configuration électronique à:
Cu: 29p, 29e^– → 1s²2s²2p⁶3s²3p⁶4s¹3d¹⁰. Les différentes entraînement le à l'extérieur environnements orbitaux (d¹⁰) il vient la raison d'autres réflexions le énergies en orbite. Cependant, de telles réflexions / réflexions ne sont que l'objet de l'ETH.

October 16, 2026 at 4:39 am in reply to:

Aus Salzen Ionen bilden (Salze-25)

Gegeben seien folgende Salze. Welches sind die zugehörenden Ionen ? Deren Anzahl muss aber nicht bestimmt werden. Das erste Beispiel ist schon gelöst.

NaCl: Na⁺ / Cl^–
Al₂O₃
MgO
KBr
FeO
Fe₂O₃
Na₂O
CaO
(NH₄)₂HPO₄
Ca(H₂PO₄)₂
Fe(NO₃)₃
Ba(ClO₃)₂
Ca(OH)₂
Ca₃(PO₄)₂
Zn(HCO₃)₂
LiCO₃
Na₂H₂P₂O₇

#3411

Lösung:

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

October 16, 2026 at 4:39 am in reply to:

Molar volume 1 (st-83)

What do you mean by molar volume?

#9109

Solution:

At zero degree centigrade and normal pressure (101'325 N/m²⁾ claimed each ideal [translation unavailable] a volumes of 0.0224 m³ or 22.4 litres.
However, this value can also be calculated:
   · V = n·R·T/p with …
   · n=1 mol, R=8.314J/(molK),
   · T=273K, p=101300N/m² is given
   · Itself a volumes of 0.0224 m³ or 22.4 of which:.

October 16, 2026 at 4:39 am in reply to:

Unterschiedliche Salze bilden (Salze-24)

Bilde folgende Salze aus den neutralen Elementen und schreibe die ausgeglichene Reaktionsgleichung auf:

NaCl
MgO
KBr
Al₂O₃
Eisen(II)oxid
Eisen(III)oxid
Natriumoxid
Calciumoxid

#3409

Lösung:

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

October 16, 2026 at 4:39 am in reply to:

Gas equation 2 (st-81)

[translation unavailable] he speaks one in fact, of the 'ideal' gas equation?

#9107

Solution:

It is not a bad assumption that the host particles move independently of each other and that the specific volume of the gas is negligibly small compared to the volume claimed. At normal conditions (zero degree centigrade and pressure on sea level) is given itself a volumes of 22.4 litres or 0.0224 m³

October 16, 2026 at 4:39 am in reply to:

Presumibilmente stesso molecole, ma diversi massa, struttura atomica (Atom-13)

La reazione di bromo con idrogeno produce il gas bromo idrogeno (HBr). I seguenti isotopi partecipano alla reazione: ¹H, ²H, ⁷⁹Br e ⁸¹Br.

Qual è l'equazione di reazione equilibrata?
Quante diverse molecole di HBr di massa diversa si formano?
Quale combinazione di isotopi sarebbe la HBr più leggera, quale è la molecola di HBr più pesante? Indicazione inclusa massa molare.

Supposizioni per la masse molari: si calcola con la semplificazione che m(Proton) = m(Neutron). si applica quanto segue:
M(¹H) = 1 g/mol, M(²H) = 2 g/mol, M(⁷⁹Br) = 79 g/mol e M(⁸¹Br) = 81 g/mol.

#7143

Soluzione:

H₂ + Br₂ ⇄ 2 · HBr
C’è 4 diverse possibili coppie:
¹H-⁷⁹Br, ¹H-⁸¹Br, ²H-⁷⁹Br e ²H-⁸¹Br
M(¹H-⁷⁹Br) = 80 g/mol, M(¹H-⁸¹Br) = 82 g/mol, M(²H-⁷⁹Br) = 81 g/mol, M(²H-⁸¹Br) = 83 g/mol
Facile molecola ¹H-⁷⁹Br, più pesante molecola ²H-⁸¹Br

October 16, 2026 at 4:39 am in reply to:

Dessinez le modèle atomique de Bohr (Atom-34)

Dessinez le modèle atomique de Bohr pour les éléments suivants: fluor, magnésium, silicium.
Pour chaque modèle, les informations suivantes doivent également être fournies:
– nombre électrons de valence
– Indication de la coque de valence
– les écailles qui appartiennent au tronc nucléaire
– Chargement de la corbeille

#5177

Solution:

encore à faire

October 16, 2026 at 4:39 am in reply to:

Generale, struttura atomica, isotopo (Atom-12)

Che cosa hanno in comune i tre atomi di silicio ²⁸Si, ²⁹Si, ³⁰Si, cosa è diverso?

#7141

Soluzione:

I protoni e gli elettroni sono sempre uguali, ma il numero di neutroni è diverso. Le proprietà chimiche dei tre isotopi sono uguali, solo che il nucleo è pesante diversamente.

October 16, 2026 at 4:39 am in reply to:

Modèle atomique de Bohr (Atom-33)

Pour tous les éléments suivants, indiquez combien

Les électrons qu'ils ont au total
Combien de électrons de valence sont présents

Les atomes: carbone, magnésium, phosphore, argone

#5175

Solution:

C: 6 e^– au total, couches n=1: 2e^–, n=2: 4e^–
électrons de valence: 4
Mg: 12 e^– au total, couches n=1: 2e^–, n=2: 8e^–, n=3: 2e^–
électrons de valence: 2
P: 15 e^– au total, couches n=1: 2e^–, n=2: 8e^–, n=3: 5e^–
électrons de valence: 5
Ar: 18 e^– au total, couches n=1: 2e^–, n=2: 8e^–, n=3: 8e^–
électrons de valence: 8

October 16, 2026 at 4:39 am in reply to:

Ionen, Salzformel aus Salzname herleiten (Salze-23)

Gegeben seien die Namen folgender Salze. Ergänze die Tabelle!

Name des Salzes	Kation	Lewisformel des Anions	Salzformel
Natriumsulfat
Kaliumnitrat
Calciumphosphat
Magnesiumcarbonat
Ammoniumchlorid
Blei(II)-sulfid
Eisen(II)-oxid
Eisen(III)-oxid
Aluminiumcarbid
Kaliumpermanganat
Natriumacetat
Aluminiumoxid

#3407

Lösung:

Name des Salzes	Kation	Lewisformel des Anions	Salzformel
Natriumsulfat	Na⁺	SO2-4	Na₂SO₄
Kaliumnitrat	K⁺	NO–3	KNO₃
Calciumphosphat	Ca²⁺	PO3-4	Ca₃(PO4)₂
Magnesiumcarbonat	Mg²⁺	CO2-3	MgCO₃
Ammoniumchlorid	NH+4	Cl^–	NH₄Cl
Blei(II)-sulfid	Pb²⁺	S^2-	PbS
Eisen(II)-oxid	Fe²⁺	O^2-	FeO
Eisen(III)-oxid	Fe³⁺	O^2-	Fe₂O₃
Aluminiumcarbid	Al³⁺	C^4-	Al₄C₃
Kaliumpermanganat	K⁺	MnO–4	KMnO₄
Natriumacetat	Na⁺	CH₃COO^–	NaCH₃COO
Aluminiumoxid	Al³⁺	O^2-	Al₂O₃

October 16, 2026 at 4:39 am in reply to:

Gas equation 1 (st-80)

What is the general (ideal) gas equation and what is it?

#9105

Solution:

It is p·V = n·R·T, where …
   · p the pressure (in N/m²⁾
   · T the temperature (in kelvins)
   · n the number Mol
   · V the volume (in m³⁾
   · and R is a constant (8.314 J·mol·K).

October 16, 2026 at 4:39 am in reply to:

Calculate using the reaction equation 7 (st-78)

Calculate the masses of all the particles involved in the following reaction equations

PbO + C → Pb + CO₂, given 0.528 g C.
M(PbO) = 223.2 g/mol, M(Pb) = 207.2 g/mol, M(CO₂) = 44 g/mol
NaCl → Na + Cl₂, given be 100 g Na.
M(NaCl) = 58.5 g/mol, M(Cl₂) = 71 g/mol
Fe + O₂ → Fe₂O₃, give 100 tonnes of iron.
M(O₂) = 32 g/mol, M(Fe₂O₃) = 159.6 g/mol
Sodium (Na) and chlorine gas (Cl₂) are produced by melting electrolysis (a special technical procedure) from sodium chloride (NaCl). Write the reaction equation and calculate how much gram of sodium chloride is required to obtain 100.0 g of sodium.
Calcium carbide (CaC₂) is a solid that forms a flammable gas with water.
reaction equation: Ca + 2·C → CaC₂
How much kg of CaC₂ do you get when you use 50 kg of Ca? How much kg of carbon is needed for this?

#9103

Solution:

· The following formulae shall apply:
n=m/M or m = n·M
· See table below
(a) Note that the equation still needs to be balanced:
   2·PbO + C → 2·Pb + CO₂
   the masses sought are:
   · m(PbO) = 19.64 g
   · m(Pb) = 18.23 g
   · m(CO₂) = 1.94 g
(b) Note that the equation still needs to be balanced:
   2·NaCl → 2·Na + Cl₂
   the masses sought are:
   · m(NaCl) = 254.4 g
   · m(Cl₂) = 154.4 g
(c) Note that the equation still needs to be balanced:
   4·Fe + 3·O₂ → 2·Fe₂O₃
   Notice that there are 100 tons of iron. The bill is in g/mol.
   m(O₂) = about 42.9 tonnes
   m(Fe₂O₃) = approximately 142.7 tonnes
(d) Identical with exercise (b) reaction equation:
2·NaCl → 2·Na + Cl₂
(e) the masses sought are:
   · Note that the figures are in kg. Conversion to g
   · m(CaC₂) = 80 kg
   · m(C) = 30 kg

substance	M(g/mol)	m(g)	n(mol)
a) PbO	223.2	19.64	2·0.044=0.088
a) C	12	0.528	0.528/12 = 0.044
a) Pb	207.2	18.23	2·0.044 = 0.088
a) CO₂	44	1.94	1·0.044 = 0.044
–––	–––	–––	–––
b) NaCl	58.5	254.4	4.35
b) Na	23	100	100/23 = 4.35
b) Cl₂	71	154.4	4.35/2 = 2.174
–––	–––	–––	–––
c) Fe	55.9	10⁸	10⁸/55.9 = 1'788'908
c) O₂	32	42'933'810.4	1'788'908/4·3= 1'341'681
c) Fe₂O₃	159.6	142'754'919.5	1'788'908/2= 894'454
–––	–––	–––	–––
d) NaCl	58.5	254.4	4.35
d) Na	23	100	100/23 = 4.35
d) Cl₂	71	154.4	4.35/2 = 2.174
–––	–––	–––	–––
e) Ca	40	50'000	50'000/4=1250
e) C	12	30'000	2·1250=2500
e) CaC₂	64	80'000	1·1250=1250

October 16, 2026 at 4:39 am in reply to:

Elementi diversi di massa uguale (Atom-11)

Inserimento{isotopi:} notare due diverse elementi (inclusi numero di massa), quali le stesso masse hanno.
Accettazione: m(p) = m (n), la massa il elettroni dovrebbe in questa esercizio trascurato sono.

#7139

Soluzione:

La stessa massa significa quindi, in definitiva, che la numero di massa sarebbe uguale. C’è centinaia diversi soluzioni, z.B.:

m(¹³C) = m(¹³N)
m(¹³C) = m(¹³O)
m(⁴He) = m(⁴Li)

Si noti che …: gli elementi sono diversi (numero p diverso)

October 16, 2026 at 4:39 am in reply to:

Nommage des électrons (Atom-32)

Comment s'appellent les électrons dans la dernière / dernière écaille ?

#5173

Solution:

Ce sont les électrons de valence.

October 16, 2026 at 4:39 am in reply to:

Bilde Salze aus Ionen, 2 (Salze-22)

Welche Salze ('Salzformeln') lassen sich aus den folgenden Ionen bilden:
Al³⁺ / NH+4 / Fe²⁺ / Fe³⁺ / Na⁺ / Mn²⁺ / O^2- / S^2- / NO₃^– / Br^– / SO2-4 / H₂PO₄^– / CO2-3 / PO3-4
Hinweis: die Ionen sollen nach ihrer Ladung aufgelistet werden, also zuerst die einfach positiven, dann die zweifach positiven etc. Und: wie auch sonst: mit einer Tabelle wird die Sache übersichtlicher.

#3405

Lösung:

	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
O^2-	Na₂O	(NH₄)₂O	FeO	MnO	Al₂O₃	Fe₂O₃
S^2-	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

October 16, 2026 at 4:39 am in reply to:

Bilde Salze aus Ionen, 1 (Salze-21)

Welche Salze ('Salzformeln') lassen sich aus den folgenden Ionen bilden:
Al³⁺ / NH+4 / Fe²⁺ / Fe³⁺ / Na⁺ / Mn²⁺ / O^2- / S^2- / NO₃^– / Br^– / SO2-4 / H₂PO₄^– / CO2-3 / PO3-4

#3403

Lösung:

	Al³⁺	NH+4	Fe²⁺	Fe³⁺	Na⁺	Mn²⁺
O^2-	Al₂O₃	(NH₄)₂O	FeO	Fe₂O₃	Na₂O	MnO
Br^–	AlBr₃	NH₄Br	FeBr₂	FeBr₃	NaBr	MnBr₂
S^2-	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

October 16, 2026 at 4:39 am in reply to:

Calculate using the reaction equation 6

(st-77)

A car consumed on 100 km 10 litres [translation unavailable].
(a) Calculate the mass of the carbon dioxide produced.
(b) Calculate the volume taken up by the resulting carbon dioxide.
Assumption: Diesel consists of a single component: C₁₅H₃₂

#9101

Solution:

1. Note: it holds to note, that the density of the C₁₅H₃₂ to determine would be. A very short Internet search gives a value of about 0.75 g/ml. The indication makes sense, diesel is floating on the water.
2. Note: the expression '10 litres [translation unavailable] are consumed' holds it of this kind to interpreting, that the [translation unavailable] Accordingly, the (balanced) reaction equation must be drawn up. How commonly holds, that at combustion reaction – where: possible – water and carbon dioxide to be created.

a) C₁₅H₃₂ (l) + 23·O₂ (g) → 15·CO₂ (g) + 16·H₂O (l)
With Dichte(C₁₅H₃₂) = 0.75 g/ ml to be given itself at 10 litres: m(C₁₅H₃₂)=7.5 kg

substance	M(g/mol)	m(g)	n(mol)
C₁₅H₃₂	212	7500	7500/212 = 35.4
CO₂	44	23349	15·35.4 = 530.6

The accounts shows, that approx. 23 kg CO₂ to be created. Assuming that about 100 km can be driven with 10 litres of diesel, the result is a CO₂ excluded of approx. 230 grams. The result corresponds quite well to the current values that modern engines can produce.

b) note: it holds at normal conditions, that any of these [translation unavailable] (thus also carbon dioxide) per mol 22.4 litres claimed.
Thus would be the volumes 530.6 · 22.4 = 11885 litres or Approximately 11.8 m³ or A dishes with a length of pages of approx. 2.3m, fully filled with CO₂.

Videoantwort:
Erklärung in Form eines Videos, Dauer 9:06

Direkter Link zum Fullscreen-Modus hier

Mitschrift:
Transcript of the video as a PDF [hier]

October 16, 2026 at 4:39 am in reply to:

Che cosa è un isotopo (Atom-10)

Che cosa è un isotopo?

#7137

Soluzione:

Gli isotopi (un elemento) hanno lo stesso numero di protoni, ma il numero di neutroni è diverso.

October 16, 2026 at 4:39 am in reply to:

Électrons de valence (Atom-31)

Quels sont les électrons de valence ?

#5171

Solution:

Les électrons de valence sont ceux qui électrons, quelles se dans le à l'extérieur écailles se trouvent. Ces nombre peut de au système de périodes dérivées sont, par le z.B. De à gauche après à droite compté est. Ainsi désigne z.B. Nitrogen 5 électrons de valence sur. Au total a il par ailleurs,, là les numéro atomique 7 est, 7 protons et donc dans le neutres état 7 électrons. Mais ce nombre total d'électrons ne doit pas être confondu avec les électrons qui ne se trouvent que dans la coquille la plus extrême: les électrons de valence.

October 16, 2026 at 4:39 am in reply to:

Salzformel bilden (Salze-20)

Schreibe alle (binären) Salzformeln auf, welche sich aus den häufigsten Ionen folgender Elemente bilden lassen:
Brom, Sauerstoff, Kalium, Aluminium, Calcium sowie Stickstoff

#3401

Lösung:

Die häufigsten Ionen wären Br^–, O^2-, K⁺ , Al³⁺ , Ca²⁺ , N^3-