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October 16, 2026 at 4:43 am in reply to:

Nomenclatura, disegno delle molecole, 2 (oc-11)

I seguenti nomi di varie molecole sono stati indicati. Disegnare la molecola di cui si tratta.

esano
2 metil-octano
3,4 -Di- Ethyl-2,2, 5 -Tri-metil-decan
Ethanolo (anche Ethan-1-ol)
Propan-1,2, 3 – tri-ol
Butan-1-[traduzione mancante]
Butan-2-su
Ciclohexanolo
4-metillio-Pentan-1-[traduzione mancante]
2-metillio-Pentan-2-[traduzione mancante]
3 – Amino- Heptan-1 -ol
Butanal di 3-idrossi
2,3-dimetil-Butan-2,3-diolo
2,3-dimetil pentano
1-Ethyl-1-Methyl cyclobutan
3 – Ethyl-2,2, 4 – Tri-metil- esano

#7665

Soluzione:

{a) esano }	/\/\/
{b) 2-metil-octano: }	@:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; @n(8,90)/@n(7,90)\@n(6,90)/@n(5,90)\@n(4,90)/@n(3,90)\@n(2,-90)<\|>/@n(1,90)
{c) 3,4 – Di-etil- 2,2, 5 – Tri-metil-decan: }	/<_q4><_p4>/<`\|/>\<\|\>/<`\|>\/\/\
{d) Ethanolo: }	/\OH
{e) Propan-1,2, 3 – tri-ol:}	HO/\<\|OH>/\OH
{f) Butan-1-[traduzione mancante]: }	/\/`\|\|O
{g) Butan-2-su:}	@:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; @n(4,90)/@n(3,90)\@n(2,-90)<\|\|O>/@n(1,90)
{h) Ciclohexanolo}	`/`-`\/-\-OH
{i) 4-metillio-Pentan-1-[traduzione mancante]}	\<\|>/\/\OH { } { o: } HO/\/\<\|>/
{j) 2-metillio-Pentan-2-[traduzione mancante]}	HO/<_q4><_p4>/\/
{k) 3 -amino- Heptan-1 -ol}	HO\/\<\|NH2>/\/\
{l) 3-idroxy butanole}	\<\|OH>/\//O
{m) 2,3-dimetil-butane-2,3-diolo }	HO-<_q4><_p4>–<_q4><_p4>-OH
{n) 2,3-dimetil pentano }	/<`\|>\<\|>/\
{o) 1 – Ethyl-1 -Methyl cyclobutan }	(`/)(`\/)\|-`\|_#1
{p) 3 – Ethyl-2,2, 4 -Tri-metil- esano }	@:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; @n(1,90)/@n(2,90)<_(A225)><_(A-45)>\@n(3,-90)<\|\>/@n(4,90)<`\|>\@n(5,90)/@n(6,90)

October 16, 2026 at 4:43 am in reply to:

Calculate various values (pH etc) (SB-44)

Calculate / add the values in the following table.

c(H₃O⁺ mol/l)	c(OH^– mol/l)	pH	pOH
0.001
0.01
	0.1
	0.2
		12
			11

#9627

Solution:

As a reminder: the ionic product applies to each concentration, for each pH or pOH respectively.
ionic product: c(H₃O⁺)·c(OH^–) = 10^-14 mol²/l²

c(H₃O⁺ mol/l)	c(OH^– mol/l)	pH	pOH
[translation unavailable]{0.001 = 10^-3}	10^-11	3	11
[translation unavailable]{0.01 = 10^-2}	10^-12	2	12
10^-13	[translation unavailable]{0.1 (10^-1)}	13	1
10^-14/0.2 = 5·10^-14	[translation unavailable]{0.2 }	13.3	0.7
10^-12	10^-2= 0.01	[translation unavailable]{12}	2
10^-3 = 0.001	10^-11	3	[translation unavailable]{11}

October 16, 2026 at 4:43 am in reply to:

Nomenclatura, disegno delle molecole, 1 (oc-10)

Ci sono 6 molecole, una delle quali è chiaramente diversa da tutte le altre, quale e perché?

{e) } \/\|`/<`\>|<`/>\

{f) } \|`/|<`/>\/<`|>\

#7663

Soluzione:

L’ultima molecola (f) è diversa da tutte le altre. (a) e) sono identiche, anche se hanno un aspetto diverso.
La nomenclatura sistematica aiuta chiaramente a evidenziare le differenze. Tenere presente che la numerazione degli atomi di carbonio non deve necessariamente essere effettuata da sinistra a destra, ma in modo che la somma dei rami sia il più piccola possibile.

{a) } @:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; @n(1,-70)\@n(2,50)<\|>/@n(3,-30)<`\|>\@n(4,50)/@n(5,-50)\@n(6,50)/@n(7,-50)\@n(8,50)		2,3-di-metil-octano
{b) } @:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; @n(1,80)/@n(2,90)<`\|>\@n(3,-90)<\|>/@n(4,90)\@n(5,90)/@n(6,-90)\@n(7,45)\|@n(8,90)		2,3-di-metil-octano
{c) } @:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; @n(8,90)\@n(7,90)/@n(6,90)\@n(5,90)/@n(4,90)\@n(3,-90)<\|@n(2,90)<`/@n(1,90)>\>/		2,3-di-metil-octano
{d) } @:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; /<`\|@n(1,-90)>@n(2,90)\@n(3,-90)<\|>/@n(4,90)\@n(5,90)/@n(6,90)\@n(7,90)/@n(8,90)		2,3-di-metil-octano
{e) } @:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; @n(8,90)\@n(7,90)/@n(6,90)\@n(5,-90)\|@n(4,90)`/@n(3,-90)<`\>\|@n(2,90)<`/@n(1,90)>\		2,3-di-metil-octano
{f) } @:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; @n(8,-90)\@n(7,-90)\|@n(6,90)`/@n(5,-90)\|<`/>@n(4,90)\@n(3,90)/@n(2,90)<`\|@n(1,-90)>\		2,4-di-metil-octano

October 16, 2026 at 4:43 am in reply to:

Disegni di alcoli diversi, 2 (oc-05)

Ogni molecola da disegnare deve avere quattro atomi di carbonio.
…. Alcol primario, alcol secondario e alcol terziario.

#7661

Soluzione:

{alcol primario: } /\/\OH

{alcol secondario: } /<`|OH>\/

{alcol terziario: } –<|><`|>-OH

October 16, 2026 at 4:43 am in reply to:

_siehe_auch_099 (nolink-099)

espace réservé

#5695

Solution:

espace réservé

October 16, 2026 at 4:43 am in reply to:

pH value 1 (SB-43)

Calculate the pH of a 3-molar acid solution.

#9625

Solution:

The pH value is the negative (decade) logarithm of the concentration of H₃O⁺ particles in a solution. The concentration must be in the range of 1 mol/l to 10^-14 mol/l. Thus a possible pH value from pH = -log 1 = 0 to pH = -log 10^-14 = 14 is obtained.
The answer (to the question) is thus: it cannot be said, since the concentration is greater than the concentration of as 1 mol per litre.

October 16, 2026 at 4:43 am in reply to:

Logarithm (SB-42)

Why is the logarithm used?

#9623

Solution:

At calculations round to the subjects acid-base chemistry has one it very often with very small concentrations to doing, this corresponds to also the laboratory everyday life. Usually, the concentrations move somewhere in the millimeter range. Are well, you know, additionally the concentrations the H₃O⁺ considered, so are these often still small, z.B. 0.00001 mol/l or 0.000001 mol/l. Which of the two concentrations is less? And you may have made the mistake of counting the zeros… tediously. If we take the logarithm and multiply it by -1, we get a size that is much more intuitive even in everyday laboratory life.

October 16, 2026 at 4:43 am in reply to:

_siehe_auch_003 (nolink-003)

[traduction manquante]

#5693

Solution:

[traduction manquante]

October 16, 2026 at 4:43 am in reply to:

Disegnare diversi tipi di alcol (oc-04)

Disegnate tre alcoli diversi:

alcol primario
alcol secondario
alcol terziario

#7659

Soluzione:

/\OH { } \<|>/\OH { } \/<`|>\/OH
/\<|HO>/ { } \/<`|OH>\| { } /<`|>\/<`|OH>\/
OH|_(A75);/#2\ { } OH|_(A75);/#2\/`| { } OH|_(A75);/#2\/\|

October 16, 2026 at 4:43 am in reply to:

Disegnare diversi gruppi funzionali, 3 (oc-03)

Disegnare ogni molecola che contemporaneamente contenga tutti i gruppi funzionali, la struttura di base deve essere sempre una struttura di alghe

Alchene, alcol e acido carbossilico
Alchene, aldeide, chetone e ammide
Acido carbossilico, alcol e chetone
Amin, ammide, acido carbossilico e Aldeyhd

#7657

Soluzione:

/<`||>\<|OH>/<`||O>\OH
H/<`||O>\/\\/<`||O>\/\<||O>/NH2
HO/<`||O>\<|OH>/<`||O>\
H2N/\N<_(y.5)H>/<`||O>\<|<`//O>\OH>/<`||O>

October 16, 2026 at 4:43 am in reply to:

Definition of pH (SB-41)

What's the pH?

#9621

Solution:

Basically a concentration ('number of particles per volume'). At pH the H⁺ (or more precisely H3O⁺) are considered. Since this number ranges from 1 mol/l to 0.00000000000001 mol/l, the Overview of the (negative-tooth) logarithm, specifically: pH = -log(c[H₃O⁺]).
So that is given itself a tensile width of the pH's of -log(1) = -log(10⁰)= 0 to to -log(0.00000000000001) = log(10^-14)= 14

October 16, 2026 at 4:43 am in reply to:

_siehe_auch_002 (nolink-002)

[traduction manquante]

#5691

Solution:

[traduction manquante]

October 16, 2026 at 4:43 am in reply to:

Disegnare diversi gruppi funzionali, 2 (oc-02)

Disegnare la formula strutturale per ciascuna esercizio. Al di là del gruppo funzionale non possono essere disegnati altri gruppi funzionali, sono ammessi solo collegamenti carbonio-idrogeno

Alcol lineare, non ramificato, con esattamente 3 atomi di carbonio
Un alcol ramificato con esattamente 5 atomi di carbonio
Tutti gli alcoli (branchiati, non ramificati) con esattamente 4 atomi di carbonio
Tre diversi etere con un totale di 4 atomi di carbonio
Il più piccolo possibile (quanti più C possibile) chetone
Aldeide minimo possibile (quanti più C possibile)
Il minimo possibile (qualsiasi C possibile) Amino
Il minori possibili (così pochi C come fattibile) estere

#7655

Soluzione:

{a) }		/\/OH
{b) }		/\<\|>/\OH
{c) }		/\/\OH { } /<`\|>\/OH { } OH\|_(A75);/#2\
{d) }		/O\/\ { } /\O/\ { } -_p_pO_p_p
{e) Il più piccolo chetone }		/<`\|\|O>\
{f) Aldeide minimo}		H/<`\|\|O>\H { o: } <`\|\|O>
{g) Minime amine}		H2N/
{h) il più piccolo estere}		<`\|O\|>\O/

October 16, 2026 at 4:43 am in reply to:

Production of a specific concentration (SB-40)

Describe the manufacture of the following solutions in the laboratory: the solvent is always distilled water.
Adopted the substances be as salts of which: and solving itself fully in water.

(a) 1 litre of 0.3 M NaOH solution
(b) 1 litre of 0.4 M HCL solution
(c) 25 ml 0.5 M NaOH solution
d) 30 ml 0.6 M H₂SO₄

#9619

Solution:

The abbreviation 'M' is a note on the concentration, so I mean, z.B. 0.3 M, that 0.3 mol the the following: substance dissolved in one liter of water, in other words: 0.3 mol/l
With c=n/V and n=m/M follows: c=m/M/V respectively m=c·V·M
Respect: here is well, you know, the 'M' the molar mass ! Depending on the situation , the 'M' stands for concentration or molar mass !

Action to be taken to the manufacture the solution: the solids, z.B. NaOH is weighted, in a z.B. Glass of glass given and so much distilled water as desired. Note that the solid requires a small volume of its own, but this need not be taken into account.

a) m(NaOH)=c(NaOH)·V·M(NaOH) = 0.3 mol/l · 1.0 l · 58.5 g/mol = 17.55 g
Thus, 17.55 g of NaOH is weighed and the glass is filled with distilled water to the mark of 1.0 litres.
b) m(HCl)=c(HCl)·V·M(HCl) = 0.4 mol/l· 1.0 l · 36.5 g/mol = 14.6 g
So you have to weigh 14.6 grams of HCl and fill the glass with distilled water until you mark 1.0 liters.
c) m(NaOH)=c(NaOH)·V·M(NaOH) = 0.5 mol/l · 0.025 l · 58.5 g/mol = 0.73 g
Weighing 0.73 g of NaOH and filling it with water until 25 ml is marked.
d) m(H₂SO₄) = c(H₂SO₄) · V · M(H₂SO₄) = 0.6 mol/l · 0.03 l · 98 g/mol = 1.764 g
Weigh 1.764 g H₂SO₄ and fill with this water until the marking of 30 ml.

October 16, 2026 at 4:43 am in reply to:

_siehe_auch_001 (nolink-001)

[traduction manquante]

#5689

Solution:

[traduction manquante]

October 16, 2026 at 4:43 am in reply to:

Disegnare diversi gruppi funzionali (oc-01)

Disegnare per ogni esercizio ciascuno di tre diversi molecole, che consistono solo di un scheletro idrocarburico (scheletro di alcano, cioè legami singoli) e del seguente gruppo funzionale:

Alcani non ramificati
Alcani ramificati
Alcani ciclici
Alchene
Alchine
Alcol
aldeide
chetone
Amino
acido carbossilico
amminoacido
estere
ammide

#7653

Soluzione:

Alcani non ramificati
\/\ ; { } ; \/\|\ ; { } ; \/\/\/\/
Alcani ramificati
/<`|>\ ; { } ; /<`|>\/ ; { } ; \<|>/\<|>/
Alcani ciclici
-_q3_q3 ; { } ; /\|`/`\`| ; { } ; _(x1.2,y1)_(x1,y-.5)_(x1,y.5)_(x1.2,y-1)_(x-1.2,y.3)_(x-1,y-.8)_(x-1,y.8)_#1
Alchene
/=/ ; { } ; /\/\// ; { } ; \//\<|>//\
Alchine
///c/\ ; { } ; -%-/ ; { } ; \\\c\<|>/\/
Alcolici:
\/\|OH ; { } ; \<|>/OH ; { } ; /\OH
Aldeide:
\||O ; { } ; O\\|\|`//`\`||/ ; { } ; H\/O`/|\|`/`\`|<_(a-30)><`->/
Cettoni:
/<`||O>\ ; { } ; /\/<`||O>\ ; { } ; /\<||O>/
[traduzione mancante]:
/\NH2 ; { } ; /<`|NH2>\/ ; { } ; /\/\/\/NH2
Acidi di carbonio:
OH`/|O`| ; { } ; Cl`/|`/N`\\<`/Cl>`|//\|\|O`|/OH ; { } ; HO/`|O|\/\/\|O`|/OH
Aminoacidi:
H2N-CH<|$color(red)CH3>-$color()COOH ; { } ; |`//`\`||/\/`|O|\N<_(y.5)H>/\|O`|/OH ; { } ; \<|>/`|<`\dH2N>/`|O|\OH
estere:
|`//`\`||/\/`|O|\O/\||\//`|`\`/ { } H|`/O/\O/\|`/O|`/O/\H { } /O\|O`|/\/`|O|\O/
[traduzione mancante]:
H2N/\O { } /\|O`|/N<_(y-.5)H>\ { } -_qNH_q<`|O|>_q_q

October 16, 2026 at 4:43 am in reply to:

La structure les prédictions, exercice 1 (oc-50)

Le résultat suivant d'une analyse élémentaire est donné: C 12.79%. H 2.15%, le reste étant le brome

Qu'est-ce que le formule moléculaire?
Dessinez le molécule en fonction du formule moléculaire. Qu'est-ce qui vous éclaire ?
Comment voir le à attendu NMR le (les deux!) les structures de?

#5687

Solution:

Remarque sur les pourcentages: il s’agit toujours de pourcentages de masse. Pour simplifier les calculs, on commence par 100 grammes de substance.

Prise de 100 grammes, soit 12.79 g de C, 2.15 g de H et 85.07 g de Brom
n(C) = 12.79/12 = 1.07 mol ≈ 1 mol
n(H) = 2.15/1 = 2.15 mol ≈ 2 mol
n(Br) = 85.07/79.9 = 1.07 mol ≈ 1 mol
{alors } CH2Br { Laisse se mais pas dessiner, règle de l’octet}
{donc} (CH2Br)2 {ou} C2H4Br2
En graphisme, il y a deux façons de calculer les atomes de brome au même C ou par atome de carbone selon un bromatone
Br\/\Br { } /<`|Br>\Br

October 16, 2026 at 4:43 am in reply to:

Interpretation of pKs value, 1 (SB-27)

The pKs(H₂SO₄) is -3, pks(HSO–4) = 1.92.
Which of the following statements is correct? And almost more importantly, why?

c(H₂SO₄) > c(HSO–4 ) > c(SO2-4 )
c(HSO–4) > c(H₂SO₄) > c(SO2-4)
c(H₂SO₄) = c(HSO–4) > c(SO2-4)
c(HSO–4) > c(SO2-4) > c(H₂SO₄)
c(SO2-4) > c(HSO–4) > c(H₂SO₄)

#9617

Solution:

The pKs value for H₂SO₄ is yes -3. This specifically means that almost all H₂SO₄ decomposes to HSO–4 and H⁺.
The pKs value of 1.92 (i.e. approximately 2) for HSO–4 states that at the start of (i.e. B) 1000 HSO–4 itself in the GW approx. A one-hundredth (pKs = 2 = -log(0.01) = -log(1/100)) on the page of the products (thus SO2-4) is located, the large remains remaining at the same time, reactant (: HSO–4).
This gives us the following picture:
c(HSO–4) > c(SO2-4) > c(H₂SO₄)

October 16, 2026 at 4:43 am in reply to:

Frequency of particles (SB-26)

What kind of particles are present in an aqueous 1 M H₂SO₄ solution? In what relative quantities are they present? Make a ranking, start with the most common particle.
pKs(H₂SO₄) is -3, pks(HSO–4) = 1.92.
Note: Es [translation unavailable] only following particles before: H₂O, H₃O⁺, HSO–4, SO2-4, (almost) no H₂SO₄, there sulphuric acid a strong acid is

#9615

Solution:

The pKs(H₂SO₄) = -3, which means that H₂SO₄ is a very strong acid, and so it releases almost 100% of its first (!) H⁺, but that leaves almost no H₂SO₄.
x > y > z > c(H₂SO₄)
The additionally 1 M H₂SO₄ it says:, that per litres water (=1000 g/M(H₂O) = 55 mol) only a Mol H₂SO₄ present is. So the water is clearly in excess: 55.5 mol H₂O + 1 mol H₂SO4 –> 54.5 mol H₂O + 0 mol H₂SO₄ + 1 mol H⁺
c(H₂O) > x > y > z > c(H₂SO₄). Open up are thus still c(H₃O⁺), c(HSO–4), c(SO2-4)
The pks(HSO–4) = 1.92, thus approx. 2 it says:, that at the same time, starting of (z.B.) 1000 HSO–4 itself in the GW approx. A one-hundredth (pKs = 2 = -log(0.01) = -log(1/100)) on the page of the products (thus SO2-4) is located, the large remains remaining at the same time, reactant (: HSO–4). It holds u.a. Thus:
c(HSO–4) > c(SO2-4)
Open up is still the question after H₃O⁺. From the the first reaction is generated at least a Mol, at other follow-up actions to be created other H₃O⁺. All other concentrations moving itself in the size orders of which: a Mol.
The following sequence is thus obtained:
c(H₂O) > c(H₃O⁺) > c(HSO–4) > c(SO2-4) > H₂SO₄.

October 16, 2026 at 4:43 am in reply to:

Réaction des groupes fonctionnels, réactions de condensation 3 (oc-42)

Es soient suivantes produits de réaction de condensation donné. Dessinons par exemple les formes squelettiques des réactifs correspondants. Notez que la notation a été faite avec une flèche de rétrosynthèse. Faites de même.

{a1) } /\/O\/\	+	H2O	⇒	/\/OH + HO\/\
{a2) } \<\|>/\O/\<\|>/	+	H2O
{a3) } \/<`\|>\/O\/<`\|>\/	+	H2O
{a4) } /<`\|>\/<`\|O/<`\|/<`\|>\>\/>\/	+	H2O
{a5) } $C()/<_(A225)><_(A-45)>\O/<_(A225)><_(A-45)>\	+	H2O
{a6) } \<\|>/<`\|>\O/<`\|>\<\|>/	+	H2O
{b1) } /\O/	+	H2O
{b2) } /<`\|>\O/\	+	H2O
{b3) } \<\|>/<`\|>\O/\	+	H2O
{c1) } \/\O/<`\|O\|>\	+	H2O
{c2) } \<\|>/<`\|>\O/<`\|O\|>\<\|>/	+	H2O
{c3) } /\<\|O`\|>/O\/<`\|>\	+	H2O
{c4) } -_qO_q<`\|O\|>_q_q	+	H2O
{d1) } /<`\|O\|>\O/<`\|O\|>\	+	H2O
{d2) } `\|O\|\O/`\|O\|	+	H2O
{d3) } `/`\`\|/\\|\/`\|O\|\O/`\|O\|\/`\|/\\|`/`\	+	H2O
{e1) } \/`\|O\|\O/`\|O\|	+	H2O
{e2) } \/`\|O\|\O/`\|O\|\<\|>/\	+	H2O
{e3) } /`\|O\|\O/`\|O\|\<\|>/	+	H2O
{f1) } /N<_(y-.5)H>\\|O`\|/	+	H2O
{f2) } \<\|>/N<_(y-.5)H>\\|O`\|/\	+	H2O
{f3) } \<\|>/N<`\|>\\|O`\|/\	+	H2O
{f4) } -_qNH_q<`\|O\|>_q_q	+	H2O

#5685

Solution:

{a1) } /\/O\/\	+	H2O	⇒	{ } /\/OH + HO\/\
{a2) } \<\|>/\O/\<\|>/	+	H2O	⇒	{ } \<\|>/\OH + HO/\<\|>/
{a3) } \/<`\|>\/O\/<`\|>\/	+	H2O	⇒	{ } \/<`\|>\/OH + HO\/<`\|>\/
{a4) } /<`\|>\/<`\|O/<`\|/<`\|>\>\/>\/	+	H2O	⇒	{ } /<`\|>\/<`\|OH>\/ + /<`\|>\/<`\|OH>\/
{a5) } $C()/<_(A225)><_(A-45)>\O/<_(A225)><_(A-45)>\	+	H2O	⇒	{ } OH\|_(A75);/#2\ + \|_(A75);HO/#2\
{a6) } \<\|>/<`\|>\O/<`\|>\<\|>/	+	H2O	⇒	{ } 2* \<\|>/<`\|OH>\
{b1) } /\O/	+	H2O	⇒	{ } /\OH + HO/
{b2) } /<`\|>\O/\	+	H2O	⇒	{ } /<`\|OH>\ + /\OH
{b3) } \<\|>/<`\|>\O/\	+	H2O	⇒	{ } \<\|>/<`\|OH>\ + /\OH
{c1) } \/\O/<`\|O\|>\	+	H2O	⇒	{ } \/\OH + HO/<`\|O\|>\
{c2) } \<\|>/<`\|>\O/<`\|O\|>\<\|>/	+	H2O	⇒	{ } \<\|>/<`\|OH>\ + HO/<`\|O\|>\<\|>/
{c3) } /\<\|O`\|>/O\/<`\|>\	+	H2O	⇒	{ } /\<\|O`\|>/OH + \<\|>/<`\|OH>
{c4) } -_qO_q<`\|O\|>_q_q	+	H2O	⇒	{ } <\OH><`\|O\|>`/\|\\|OH
{d1) } /<`\|O\|>\O/<`\|O\|>\	+	H2O	⇒	{ } /<`\|O\|>\OH + HO/<`\|O\|>\
{d2) } `\|O\|\O/`\|O\|	+	H2O	⇒	{ } 2* $C()\|O`\|/OH
{d3) } `/`\`\|/\\|\/`\|O\|\O/`\|O\|\/`\|/\\|`/`\	+	H2O	⇒	{ } 2* `/`\`\|/\\|\/`\|O\|\OH
{e1) } \/`\|O\|\O/`\|O\|	+	H2O	⇒	{ } /\<\|O`\|>/OH + HO/`\|O\|
{e2) } \/`\|O\|\O/`\|O\|\<\|>/\	+	H2O	⇒	{ } HO\\|O`\|/<`\|>\/ + \/`\|O\|\OH
{e3) } /`\|O\|\O/`\|O\|\<\|>/	+	H2O	⇒	{ } HO\\|O`\|/<`\|>\ + /`\|O\|\OH
{f1) } /N<_(y-.5)H>\\|O`\|/	+	H2O	⇒	{ } /NH2 + HO/`\|O\|\
{f2) } \<\|>/N<_(y-.5)H>\\|O`\|/\	+	H2O	⇒	{ } \<\|>/NH2 + /\\|O`\|/OH
{f3) } \<\|>/N<`\|>\\|O`\|/\	+	H2O	⇒	{ } \<\|>/N<_(y-.5)H>\ + /\\|O`\|/OH
{f4) } -_qNH_q<`\|O\|>_q_q	+	H2O	⇒	{ } <\OH><`\|O\|>`/\|\\|NH2

October 16, 2026 at 4:43 am in reply to:

Calcolo QM 3 (qm-03)

Prezzi (calcolatori) a partire da, quanti fotoni una lampada luminosa con 100 W per seconda nel visibili area del di luce emessi.
Nota 1: Supponi, per semplificazione, che λ = 500 nm

#7651

Soluzione:

E=h·f = h·c/λ
E_total = x·h·f = x·h·c/λ
   E_total corrisponde a 100 W/s
   x è il numero di particelle a 500 nm
   x = E_total · λ / (h·c)
   x = 100J·500·10^-9 m / (6.026·10^-34 Js · 3.0·10⁸m/s)
   x = 2.5·10²⁰, cioè circa 1/1000 mol

October 16, 2026 at 4:43 am in reply to:

Statements, which ones are correct (SB-25)

It is a watered, 1 molar solution a unprotonated heavy acid manufacture ('HX'). Which of the following statements is correct? Reason for the choice.

The acid is dissolved by 0.1%
The acid is dissolved at 1%
The acid is dissolved by 10%
The acid is 100% dissolved

#9613

Solution:

The definition of a strong acid is that the acid balance is completely on the deprotonated side:

{HX} -> H^+ + {X}^- {respectively:}

{HX} + H2O -> H3O^+ + {X}^-

Where: thus no balance of which: and the acid fully decomposed, so he speaks one also of a 100% dissociation. So correct is the solution d .

October 16, 2026 at 4:43 am in reply to:

Réaction des groupes fonctionnels, réactions de condensation 2 (oc-41)

Vous allez à chaque exercice les masses suivantes: avant:
– dessiner les molécules requises avec les groupes fonctionnels
– laissez les deux molécules dessinées réagir entre elles
– en plus au produit eau se produit une nouveaux groupe fonctionnel. Quelles?
– chaque tâche partielle doit être résolue au moins trois fois
Remarque: les molécules doivent être neutres

Deux alcools identiques réagissent
Deux alcools différents réagissent
Un alcool réagit avec un acide carboxylique
Deux acides carboniques identiques réagissent entre eux
Deux acides carboniques différents réagissent entre eux
Une aminée réagit avec une acide carboxylique

#5683

Solution:

{a) Deux alcools égaux}		{éther}
{a1) } /\/OH + /\/OH	->	/\/O\/\ + H2O
{a2) } \<\|>/\OH + \<\|>/\OH	->	\<\|>/\O/\<\|>/ + H2O
{a3) } \/<`\|>\/OH + HO\/<`\|>\/	->	\/<`\|>\/O\/<`\|>\/ + H2O
{a4) } /<`\|>\/<`\|OH>\/ + /<`\|>\/<`\|OH>\/	->	/<`\|>\/<`\|O/<`\|/<`\|>\>\/>\/ + H2O
{a5) } OH\|_(A75);/#2\ + \|_(A75);HO/#2\	->	$C()/<_(A225)><_(A-45)>\O/<_(A225)><_(A-45)>\ + H2O
{a6) } 2* \<\|>/<`\|OH>\	->	\<\|>/<`\|>\O/<`\|>\<\|>/ + H2O
{b) Deux alcools différents}		{éther}
{b1) } /\OH + HO/	->	/\O/ + H2O
{b2) } /<`\|OH>\ + /\OH	->	/<`\|>\O/\ + H2O
{b3) } \<\|>/<`\|OH>\ + /\OH	->	\<\|>/<`\|>\O/\ + H2O
{c) l’alcool + acide carboxylique}		{ester}
{c1) } \/\OH + HO/<`\|O\|>\	->	\/\O/<`\|O\|>\ + H2O
{c2) } \<\|>/<`\|OH>\ + HO/<`\|O\|>\<\|>/	->	\<\|>/<`\|>\O/<`\|O\|>\<\|>/ + H2O
{c3) } /\<\|O`\|>/OH + \<\|>/<`\|OH>	->	/\<\|O`\|>/O\/<`\|>\ + H2O
{d) Deux acides carboniques égaux}		{anhydride d’acide carboxylique}
{d1) } /<`\|O\|>\OH + HO/<`\|O\|>\	->	/<`\|O\|>\O/<`\|O\|>\ + H2O
{d2) } 2* $C()\|O`\|/OH	->	`\|O\|\O/`\|O\| + H2O
{d3) } 2* `/`\`\|/\\|\/`\|O\|\OH	->	`/`\`\|/\\|\/`\|O\|\O/`\|O\|\/`\|/\\|`/`\ + H2O
{e) Deux acides carboniques différents}		{anhydride d’acide carboxylique}
{e1) } /\<\|O`\|>/OH + HO/`\|O\|	->	\/`\|O\|\O/`\|O\| + H2O
{e2) }HO\\|O`\|/<`\|>\/ + \/`\|O\|\OH	->	\/`\|O\|\O/`\|O\|\<\|>/\ + H2O
{e3) } HO\\|O`\|/<`\|>\ + /`\|O\|\OH	->	/`\|O\|\O/`\|O\|\<\|>/ + H2O
{f) amine + acide carboxylique:}		{amide}
{f1) } /NH2 + HO/`\|O\|\	->	/N<_(y-.5)H>\\|O`\|/ + H2O
{f2) } \<\|>/NH2 + /\\|O`\|/OH	->	\<\|>/N<_(y-.5)H>\\|O`\|/\ + H2O
{f3) } \<\|>/N<_(y-.5)H>\ + /\\|O`\|/OH	->	\<\|>/N<`\|>\\|O`\|/\ + H2O

October 16, 2026 at 4:43 am in reply to:

Calcolo QM 2 (qm-02)

Dopo circa 30 ore di volo, gli elettroni del vento solare raggiungono la Terra, dove possono produrre luci polari.
Prevalenti nelle luci polari sono le luci verdi, provocate dall'ispirazione dell'ossigeno a λ = 557.7 nm.
Calcolare che l'energia cinetica degli elettroni del vento solare sia sufficiente a stimolare l'ossigeno.
Nota: la distanza tra la Terra e il Sole è di 150 milioni di km
Nota 2: m(e^–)= 9.1·10^-31 kg

#7649

Soluzione:

· Idea generale: l'energia cinetica deve corrispondere almeno alla energia di eccitazione
· E_kin = ½ · m·v²
= ½ · 9.1·10^-31·(1.5·10¹¹m/(30·3600))^2 = 8.8·10^-19 J
· E_A = h·c/λ = 6.626·10^-34Js·3.0·10⁸/557.7·10^-9 = 3.6·10^-19 J
· E_kin / E_A = circa 2.4
Cioè Esso è più fabbricazione energia come energia di eccitazione presente

October 16, 2026 at 4:43 am in reply to:

Réaction des groupes fonctionnels, réactions de condensation 1 (oc-40)

Donnez les molécules suivantes. Laissez-les réagir entre elles dans une réaction de condensation.

{a) Deux alcools égaux }

{a1) } /\/OH + /\/OH

{a2) } \<|>/\OH + \<|>/\OH

{a3) } \/<`|>\/OH + HO\/<`|>\/

{a4) } /<`|>\/<`|OH>\/ + /<`|>\/<`|OH>\/

{a5) } OH|_(A75);/#2\ + |_(A75);HO/#2\

{a6) } 2* \<|>/<`|OH>\

{b) Deux alcools différents }

{b1) } /\OH + HO/

{b2) } /<`|OH>\ + /\OH

{b3) } \<|>/<`|OH>\ + /\OH

{c) l’alcool et une acide carboxylique }

{c1) } \/\OH + HO/<`|O|>\

{c2) } \<|>/<`|OH>\ + HO/<`|O|>\<|>/

{c3) } /\<|O`|>/OH + \<|>/<`|OH>

{d) deux même acide carboxylique }

{d1) } /<`|O|>\OH + HO/<`|O|>\

{d2) } 2* $C()|O`|/OH

{d3) } 2* `/`\`|/\|\/`|O|\OH

{e) Deux autres acide carboxylique }

{e1) } /\<|O`|>/OH + HO/`|O|

{e2) } HO\|O`|/<`|>\/ + \/`|O|\OH

{e3) } HO\|O`|/<`|>\ + /`|O|\OH

{f) Amine et une acide carboxylique }

{f1) } /NH2 + HO/`|O|\

{f2) } \<|>/NH2 + /\|O`|/OH

{f3) } \<|>/N<_(y-.5)H>\ + /\|O`|/OH

#5681

Solution:

solutions voir à exercice oc-41

October 16, 2026 at 4:43 am in reply to:

Acid strength (SB-22)

Given acetic acid:

CH3COOH{: } H-C<|H><`|H>-C<`||O>-O-H

Which H's are acid at the acetic acid: those who H’s at the same time, carbon or the H at the same time, oxygen?

#9611

Solution:

A connection can several H-atoms contain, z.B. Acetic acid, CH₃COOH. The at the C-atom of which: H's are less very strong polarized as the H, which at the O-atom binded is. Reason: Differences in electronegativity: C and H have similar EN values and thus have similar partial positive or negative charges. However, O and H have very different EN values, with H clearly positively partially charged, as opposed to the negatively partially charged O atom. The splitting of a H⁺ (please note:: no partly loaded but a complete [translation unavailable] loading or The failure of a complete negative loaded electrons) is thus through this large EN-difference eased. (note for the specialists: yes, I [translation unavailable], HF power difficulties, leads but at this place to far away)

October 16, 2026 at 4:43 am in reply to:

Calcolo QM 1 (qm-01)

a) Calcolare l'energia di un fotone di luce blu con una lunghezza d'onda di 450 nm.
b) Calcolare il numero di particelle luminose (fotoni) di una lampada blu con una potenza di 300 watt
Watt corrisponde a joule/secondo

#7647

Soluzione:

a) E = h·c/λ = 6.626·10^-34 Js · 299'792'458 m/s / (450·10^-9 m) = 4.4114·10^-19 J
b) E = n·h·c/λ, n = E·λ/(h·c) = 300J/s · 450·10^-9 m / (6.626·10^-34 Js · 299792458 m/s) = 6.8·10²⁰

October 16, 2026 at 4:43 am in reply to:

Rétrosynthèse, réduction (oc-35)

Les structures suivantes sont données. Comment les substances peuvent-elles être produites à partir des équipements hydrocarburants appropriés? Écrivez les étapes intermédiaires correspondantes (stables). En partie, il faut réduire (ou oxider) au bon endroit, mais ces difficultés techniques ne doivent pas être prises en compte à ce point.
Remarquez également que la notation a été faite avec une flèche de rétrosynthèse.

{a) } /\/`\|O\|\OH	⇒	/\/`\|O\|\H	⇒	/\/`\|OH	⇒	/\/
{b) } \<\|>/`\|O\|\H
{c) } \/<`\|>\\|O`\|/OH
{d) } /<`\|>\/<`\|O\|>\/
{e) } OH\|_(A75);/#2\

#5679

Solution:

{a) } /\/`\|O\|\OH	⇒	/\/`\|O\|\H	⇒	/\/`\|OH	⇒	/\/
{b) } \<\|>/`\|O\|\H	⇒	\<\|>/\OH	⇒	\<\|>/
{c) } \/<`\|>\\|O`\|/OH	⇒	\/<`\|>\\|O`\|/H	⇒	\/<`\|>\/OH	⇒	\/<`\|>\
{d) } /<`\|>\/<`\|O\|>\/	⇒	/<`\|>\/<`\|OH>\/	⇒	/<`\|>\/\/
{e) } OH\|_(A75);/#2\	⇒	/<`\|>\

October 16, 2026 at 4:43 am in reply to:

pH – pKs (SB-21)

Are pH and pKs the same?

#9609

Solution:

No. Briefly said: the pH-value corresponds to (finally) a specified number protons, the pKs-value corresponds to a readiness to pay the protons. (note: each negative decade logarithm) comparison: much money – not much money (:pH), greedy – generous (:pKs) .
Attention (!): Mathematically, however, an equation can be obtained: from the Henderson-Hasselbalch equation (pH=pKs + log …) follows purely mathematically if the log value contribution is zero, pH = pKs

October 16, 2026 at 4:43 am in reply to:

Calcolo titolazione 1 (SB-70)

Le titolazione una acido (20 ml, sconosciuti concentrazione) con una base (0.1 M) necessario fino a per il punto di equivalenza 30 ml il base. Calcola le concentrazioni dell'acido.

#7645

Soluzione:

Applicabile alla punto di equivalenza: n(Base) = n(Säure)
con c=n/V si ottiene:
c(Base)·V(Base) = c(Säure)·V(Säure)
c(Säure) = c(Base)·V(Base)/V(Säure)
c(Säure) = 0.1M · 30ml / 20 ml = 0.15 M = 0.15 mol/l

October 16, 2026 at 4:43 am in reply to:

pKs (SB-20)

What's the pK_s?

#9607

Solution:

If HX is an acid that reacts with water, the following can be defined using the chemical balance: K=(c[H₃O⁺] · c[X^–])/(c[H₂O · c[HX]) . Of the other can continued simplified are, there the water concentration approx. 1000 [translation unavailable] greater is in the comparison to a acid: K_S= (c[H₃O⁺] · c[X^–])/(c[HX]) . Instead of the K_S value, the K_W value is also spoken of.
A substance, which thus '[translation unavailable]' his protons is emitted (at water) would have thus a high K_W-value and otherwise. [translation unavailable] are the numbers small, therefore is [translation unavailable] the (negative teeth)-logarithm of which:: pK_S = -log(K_s). Quintessence: the pK_S corresponds to the readiness to pay of H⁺ (protons) a connection.

October 16, 2026 at 4:43 am in reply to:

Route d’oxydation 3 (oc-32)

Les alcools suivants doivent être présents. Des oxydants – dans le cas où: possible – tous boissons alcoolisées une fois.

{e) } `|/:a\|_#1`/`\`/<`|>|`/`\<`/HO>`|/\|0\\/`|`\0`|/\`|; #a`|<`\>/\/\<|>/

{f) } `/`-`\/-<\>\-OH

{g) } `/`-`\/-\-OH

{h) } HO\<|>/<`|NH2>\<|OH>/<`|OH>\<_(A45)><_(A-220)>/OH

{i) } HO|`/|`/`\\`|<`\<_q><_(A-70)OH>`\>//\|\O/\\/<\O/\OH>`||`\`//|

{j) } OH`\`|O|`/|`/`\\`|<`\`/HO>//\|\O|`/O/\

#5677

Solution:

{a) } /\OH	-> /`\|O\|
{b) } \<\|>/\OH	-> \<\|>/<`\|O\|>\H
{c) } \/<`\|>\/OH	-> \/<`\|>\<\|O`\|>/H {ou } \/<`\|>\\|O`\|
{d) } HO\/<`\|OH>\/OH	-> O\\/<`\|O\|>\//O
{e) } `\|/:a\\|_#1`/`\`/<`\|>\|`/`\<`/HO>`\|/\\|0\\/`\|`\0`\|/\`\|; #a`\|<`\>/\/\<\|>/	-> `\|/:a\\|_#1`/`\`/<`\|>\|`/`\<`/O/>`\|/\\|0\\/`\|`\0`\|/\`\|; #a`\|<`\>/\/\<\|>/
{f) } `/`-`\/-<\>\-OH {Aucune oxydation possible, aucun C-H}
{g) } `/`-`\/-\-OH	-> `/`-`\/-\-O`-
{h) } HO\<\|>/<`\|NH2>\<\|OH>/<`\|OH>\<_(A45)><_(A-220)>/OH	-> O\<_(A210)><\|>/<`\|NH2>\<\|O`\|>/<`\|O\|>\<_(A45)><_(A-220)>/OH
{i) } HO\|`/\|`/`\\`\|<`\<_q><_(A-70)OH>`\>//\\|\O/\\/<\O/\OH>`\|\|`\`//\|	-> O\|\|`/\|`/`\\`\|<`\<_(A130)><_(A-70)OH>`\>//\\|\O/\\/<\O/\O`\>`\|\|`\`//\|
{j) } OH`\`\|O\|`/\|`/`\\`\|<`\`/HO>//\\|\O\|`/O/\	-> OH`\`\|O\|`/\|`/`\\`\|<`\`/O/>//\\|\O\|`/O/\

October 16, 2026 at 4:43 am in reply to:

Calcoli di pH, 4 (SB-62)

Ciascuno di essi deve fornire un litro di due soluzioni A e B con la stessa pH iniziale di 4.8. A è una soluzione di acido salicylico e B una soluzione di acido acido.
Ora si aggiunge un litro di acqua a ogni soluzione.

Calcolare il nuovo pH della soluzione A e B
Qual è la soluzione in cui si verifica un maggiore cambiamento del pH?

Nota: HCl è una acido forte (pKs < 0), acido acido una acido debole (pKs > 0)

#7643

Soluzione:

Sulla base del pH, l'acido acido può essere utilizzato per l'acido forte HCl e per l'acido debole. la concentrazione è calcolata:

acido forte: HCl
pH = -log(c(H₃O⁺)), c(H₃O⁺) = 10^-pH = 0.000015 mol/l
c = n/V rispettivamente n = c · V
n = 0.000015 · 1.0 l = 0.000015 mol
Acido debole: acido acido
pH = 1/2(pKs-log(HA)), c(HA) = …
n = c·V =
Ora si aggiunge 1 litro di acqua pura a ogni soluzione, con c=n/V si ottiene:
HCl: pH = -log(c(H₃O⁺)) = -log(n/V)
pH = ha mentito

… non è ancora finito !

October 16, 2026 at 4:43 am in reply to:

Route d’oxydation 2 (oc-31)

Oxydez autant que possible les alcools suivants.

{f) } /<`|>\/<`|OH>\/

{g) } OH|_(A75);/#2\

{h) } OH|_(A75);/#2\/`|

{i) } OH|_(A75);/#2\/\|

#5675

Solution:

{a) } /\OH	->	/\O`\	->	/<\O`\>`\|OH
{b) } \<\|>/\OH	->	\<\|>/\O`\	->	\<\|>/<\O`\>`\|OH
{c) } \/<`\|>\/OH ->	->	\/<`\|>\/O`/	->	\/<`\|>\<\|OH>
{d) } /\<\|HO>/	->	/\<\|O`\|>/	->	{Aucune oxydation ultérieure possible}
{e) } \/<`\|OH>\\|	->	\/<`\|O\|>\\|	->	{Aucune oxydation ultérieure possible}
{f)} /<`\|>\/<`\|OH>\/	->	/<`\|>\/<`\|O\|>\/	->	{Aucune oxydation ultérieure possible}
{g) } OH\|_(A75);/#2\	->	{Aucune oxydation possible}
{h) } OH\|_(A75);/#2\/`\|	->	{Aucune oxydation possible}
{i) } OH\|_(A75);/#2\/\\|	->	{Aucune oxydation possible}

October 16, 2026 at 4:43 am in reply to:

Drawing various acids (SB-15)

Draw the Lewis structures of the following acids.

{(a) Acetic acid: }
{(b) Acidic acid:}
{(c) Folic acid: }
{(d) Sulphuric acid: }
{(e) Salpetic acid: }
{(e) Carbonic acid: }

#9605

Solution:

{(a) Acetic acid: }	H-C<\|H><`\|H>-C<`\|O\|>-O-H
{(b) Acidic acid:}	H-Cl
{(c) Folic acid: }	H-C<`\|O\|>-O-H
{(d) Sulphuric acid: }	H-O-S<`\|O^-><\|O^->-O-H
{(e) Salpetic acid: }	O^-\|N^+`/O/\O/H
{(e) Carbonic acid: }	HO\C\|O`\|/OH

October 16, 2026 at 4:43 am in reply to:

Calcoli di pH, miscele (SB-61)

Calcola il pH delle seguenti miscele.

Per 80 ml di una soluzione di 0.05 M di acido di sale, si devono somministrare 100 ml di una soluzione di 0.01 M di sodio.
Per 60 ml di una soluzione di 0.015 KOH, si somministrano 30 ml di una soluzione di 0.2 M di acido salicylico.
10 ml 0.2 M soluzione di acido salino sono prima di tutto [traduzione mancante] diluito e e poi con 1.0 g Ca(OH)₂ traslocato.

#7641

Soluzione:

c=n/V → n=c·V
   · 80 ml 0.05 M HCl → n(HCl)=0.004 mol
   · 100 ml 0.01 NaOH → n(NaOH)=0.001 mol
   · residuo nella reazione di neutralizzazione 0.003 mol di HCl
   · Nota: volume totale 80ml + 100 ml
   · pH=-log(0.003/0.18) = 1.78
· 60 ml 0.015 M KOH → n(KOH)=0.0009 mol
   · 30 ml 0.2 M HCl → n(HCl)=0.006mol
   · Si noti che KOH si comporta in modo simile a NaOH, quindi è anche una base
   · residuo nella reazione di neutralizzazione di 0.0051 mol di HCl
   · Nota: volume totale 60ml + 30 ml
   · pH=-log(0.0051/0.09) = 1.25
· Start- pH: pH=-log(0.2) = 0.7
   · [traduzione mancante] diluzione cioè:, che a i 10 ml ancora altri 90 ml d'altro canto. L'acqua viene fornita
   · 10 ml 0.2 M HCl → n(HCl)=0.002 mol oppure 0.002 mol
   · n(HCl)=0.002 mol → 0.002 mol H⁺
   · Volumi totali: 10ml+90ml=100 ml
   · pH=-log(0.002/0.1) = 1.7
   · Aumento per una pH-unità potenza a [traduzione mancante] diluzione significato

   · 1 g Ca(OH)₂ → n(Ca(OH)₂)=1/74.1=0.0135 mol
   · n(Ca(OH)₂)=0.0135 mol → n(OH^–)=2·0.0135= 0.027 mol
   · Reazione di neutralizzazione: 0.002 mol H e 0.027 mol OH^–
   · restano 0.025 mol OH^–
   · volume totale: 100 ml
   · pOH=-log(0.025/0.1)=0.6 → pH=14-0.6=13.4

October 16, 2026 at 4:43 am in reply to:

Route d’oxydation 1 (oc-30)

Oxydez les substances suivantes une fois par jour. Donnez à chaque oxydation ce qui ne se produit qu’à court terme Produit intermédiaire.

{a) } /\OH	->	[/<`\|OH>\OH]	->	/\O`\ + H2O
{b) } \<\|>/\OH
{c) } /\<\|HO>/
{d) } /<`\|>\/<`\|OH>\/
{e) } OH\|_(A75);/#2\

#5673

Solution:

{a) } /\OH	->	[/<`\|OH>\OH]	->	/\O`\ + H2O
{b) } \<\|>/\OH	->	[\<\|>/<`\|OH>\OH]	->	\<\|>/\\O + H2O
{c) } /\<\|HO>/	->	[/\<_(a110)HO><\OH>/]	->	/\<\|O`\|>/ + H2O
{d) } /<`\|>\/<`\|OH>\/	->	[/<`\|>\/<`\HO>\/]	->	/<`\|>\/<`\|O\|>\/ + H2O
{e) } OH\|_(A75);/#2\				{Oxydation pas possible}

! Attention ! d) contient une autre erreur, bug?! en charchem?!

October 16, 2026 at 4:43 am in reply to:

Acids in water (SB-14)

The following acids are dissolved in water. Es of which: the adopted, that the acids in full deprotonizing. Write the corresponding ions analogue to exercise (a).

{(a) Acetic acid: }	CH3COOH + H2O <=> CH3COO^- + H3O^+
{(b) Acidic acid:}
{(c) Folic acid: }
{(d) Sulphuric acid: }
{(e) Salpetic acid: }
{(e) Carbonic acid: }

#9603

Solution:

{(a) Acetic acid: }	CH3COOH + H2O <=> CH3COO^- + H3O^+
{(b) Acidic acid:}	HCl + H2O <=> Cl^- + H3O^+
{(c) Folic acid: }	HCOOH + H2O <=> HCOO^- + H3O^+
{(d) Sulphuric acid: }	H2SO4+ 2H2O <=> SO4^2- + 2H3O^+
{(e) Salpetic acid: }	HNO3+ H2O <=> NO3^- + H3O^+
{(e) Carbonic acid: }	H2CO3+ 2H2O <=> CO3^2- + 2H3O^+

October 16, 2026 at 4:43 am in reply to:

Neutralizzazioni (SB-60)

Formulare la reazioni di neutralizzazione in carattere ionico e forma la corrispondente sali, analogo alla esercizio a).

{(a) idrossido di sodio + acido salino }	Na^+ + OH^- + H3O^+ + Cl^- -> NaCl{([traduzione mancante])} + 2H2O
{b) Idroossido di sodio + acido di salpetro }
{(c) idrossido di sodio + acido di carbonio }
{(d) idroossido di potassio + acido solforico }
{(e) idrossido di potassio + acido fosforo }

#7639

Soluzione:

{(a) idrossido di sodio + acido salino }	Na^+ + OH^- + H3O^+ + Cl^- -> NaCl{([traduzione mancante])} + 2H2O
{b) Idroossido di sodio + acido di salpetro }	Na^+ + OH^- + H3O^+ + NO3^- -> NaNO3{([traduzione mancante])} + 2H2O
{(c) idrossido di sodio + acido di carbonio }	2Na^+ + 2OH^- + 2H3O^+ + 2CO3^2- -> Na2CO3{([traduzione mancante])} + 4H2O
{(d) idroossido di potassio + acido solforico }	2K^+ + 2OH^- + 2H3O^+ + SO4^2- -> K2SO4{([traduzione mancante])} + 4H2O
{(e) idrossido di potassio + acido fosforo }	3K^+ + 3OH^- + 3H3O^+ + PO4^3- -> K3PO4{([traduzione mancante])} + 4H2O

October 16, 2026 at 4:43 am in reply to:

Calcoli di pH, diluzioni (SB-49)

Calcolare i pH delle seguenti soluzioni, se il solvente è sempre acqua pura:

Una 0.03 pioggia HCl-soluzione è cento volte diluito.
Calcolare il pH prima e dopo il diluimento
20 ml 0.05 M HCl è diluito con 100 ml di acqua distillata.
Calcolare il pH prima e dopo il diluimento
Per 50 ml di 0.2 mo di soluzione di NaOH, si somministrano 100 ml di acqua distillata.
Calcolare il pH prima e dopo il diluimento

#7637

Soluzione:

Start- pH: pH=-log(0.03) = 1.5
   Un centinaio di diluzioni significa che su una soluzione ci sono 99 parti di acqua.
   z.B. 1 ml il 0.03 M soluzione più 99 ml acqua
   · 1000 ml ≙ 0.03 mol H⁺
   · 1 ml → 0.00003 mol H⁺
   · pH=-log(0.00003/0.001)=1.52
   · Si noti che 99 ml di acqua sono ammessi
   · pH=-log(0.00003/(0.001+0.099)) = 3.52
   Si noti che l'aumento di due decima potenze si è rivelato
   H⁺ concentrazione cambia se per 2 unità al pH
Start- pH: pH=-log(0.05) = 1.3
   · 20 ml di 0.05 M di HCl:
   · 1000 ml ≙ 0.05 mol H⁺
   · 20 ml → 0.001 mol H⁺
   · Si noti che 100 ml di acqua sono ammessi
   · Nuovo pH: pH=log(n/V)=-log(0.001/(0.02+0.1))= 2.079
pH di partenza: pOH=-log(0.2)=0.7, pH=14-0.7=13.3
   · 1000 ml ≙ 0.2 mol OH^–
   · 50 ml → 0.01 mol OH^–
   · Si noti che 100 ml di acqua sono ammessi
   · pOH=-log(0.01/(0.05+0.1))=1.18
   · pH=14-1.18=12.82

October 16, 2026 at 4:43 am in reply to:

Conjugated, various examples 2 (SB-13)

Enter the following reactions. Order the participating substances in corresponding acid – bases parts

{a) } NH3 + HCl <=> NH4^+ + Cl^-

{b) } HSO4^- + CN^- <=> HCN + SO4^2-

{c) } H3O^+ + HS^- <=> H2S + H2O

{d) } N2H4 + HSO4^- <=> N2H5^+ + SO4^2-

{e) } H2O + NH2^- <=> NH3 + OH^-

#9601

Solution:

Note: conjugate and corresponding SB pairs refer to the same situation

	acid 1	base 2	base 1	acid 2
a) NH₃ + HCl ⇄ NH+4 + Cl^–	HCl	NH₃	Cl^–	NH+4
b) HSO–4 + CN^– ⇄ HCN + SO2-4	HSO–4	CN^–	SO2-4	HCN
c) H₃O⁺ + HS^– ⇄ H₂S + H₂O	H₃O⁺	HS^–	H₂O	H₂S
d) N₂H₄ + HSO–4 ⇄ N₂H+5 + SO2-4	HSO–4	N₂H₄	SO2-4	N₂H+5
e) H₂O + NH–2 ⇄ NH₃ + OH^–	H₂O	NH–2	OH^–	NH₃

October 16, 2026 at 4:43 am in reply to:

Nomenclature, chiralité, 2 (oc-21)

Dessinez trois molécules, chacune ayant au moins un centre de chiralité. Les molécules dessinées doivent être neutres. Marquez les centres de chiralité avec une étoile.
(a) tous les éléments sont autorisés
Les éléments C, H et Br doivent apparaître au moins une fois dans chaque molécule; d'autres éléments ne sont pas autorisés.

#5671

Solution:

{a) } @:n(n,a,L:.2)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; \<|>/\<@n(*,-90)|NH2>/`|O|\OH

{a) } @:n(n,a,L:.2)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; HO\/<_(A-120)><_(A-60)>\<@n(*,-90)|OH>/`|O|\N<_(y.5)H>/\/`|O|\OH

{a) } @:n(n,a,L:.2)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; |`/|O`|`\`|`\\`/||\//`|0/O\@n(*,-90)/\\/`||`\`//|

{b) } @:n(n,a,L:.2)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; Br\@n(*,-90)<|>/\/

{b) } @:n(n,a,L:.2)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; //<`|Br>\<@n(*,-90)/>|`/`\`|

{b) } @:n(n,a,L:.3)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; -%-@n(*, 20)<\`/>/Br

October 16, 2026 at 4:43 am in reply to:

Conjugated, various examples 1 (SB-12)

Complete the following table with the first example solved.
Should be it for a substance z.B. No conjoined base to give (z.B. F^-), and then should the cell contents are marked with an 'X'.

	Other articles of clothing. Base	Other articles of clothing. Acid
H₂O	OH^–	H₃O⁺
HS^–
F^–
H₂PO–4
PO-34
H₃PO4
NH₃
NO–2

#9599

Solution:

	Other articles of clothing. Base	Other articles of clothing. Acid
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 4:43 am in reply to:

Nomenclature, chiralité (oc-20)

Les molécules suivantes doivent être indiquées: si la molécule contient un ou plusieurs centres de chiralité, marquez-les avec une étoile; s’il n’y a pas de centres de chiralité, indiquez kC.

{a: } /<`|>\/\/

{b: } /<`|>\<|>/<`|>\

{c: } /\<|>/\/\

{d: } /\<_(A150)><|OH>/\

{e: } Br\<|>/\/

{f: } OH`\`|O|`/|`/`\`|//\|\O|`/O/\

{g: } –<_(a30,N2)O>_q3O_q3<>_(a30)

{h: } OH|`/|<\OH>`/<|OH>`\<`/HO>`|`\O\/O\

{i: } `/O`\`/O/`|//\|\/`||`\`//

{j: } ||`/`\<`/HO>`|<`\HO>//\/<_(A-90)OH>\/N<_(y-.5)H>\CH3

{j: } \|`//<|>`\`/O/`|/`|`\c\/

{j: } `|/:a\|_#1`/`\`/<`|>|`/`\<`/HO>`|/\|0\/`|`\0`|/\`|; #a`|<`\>/\/\<|>/

#5669

Solution:

{a: } /<`\|>\/\/	{à l’échelle nationale}
{b: } /<`\|>\<\|>/<`\|>\	{l’équipement}
{c: } @:n(n,a,L:.2)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; /\@n(*,-90)<\|>/\/\	{}
{d: } /\<_(A150)><\|OH>/\	{à l’échelle nationale}
{e: } @:n(n,a,L:.2)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; Br\@n(*,-90)<\|>/\/	{}
{f: } OH`\`\|O\|`/\|`/`\`\|//\\|\O\|`/O/\	{à l’échelle nationale}
{g: } @:n(n,a,L:.2)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; @n(*,-110)-<_(a30,N2)O>_q3O_q3<>_(a30)	{}
{h: } @:n(n,a,L:.2)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; OH\|`/\|<@n(,0)\OH>`/<@n(,-90)\|OH>`\@n()<`/HO>`\|`\O\/O\@n(,-60)	{}
{i: } @:n(n,a,L:.2)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; `/O`\`/O/`\|//\\|@n(*)\/`\|\|`\`//	{}
{j} @:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; \|\|`/`\<`/HO>`\|<`\HO>//\/<>@n(*,90)<_(A-90,w)OH>\/N<_(y-.5)H>\CH3	{}
{j} @:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; @n(*,90)\\|`//<\|>`\`/O/`\|/`\|`\c\/	{}
{k} @:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; @:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; `\|@n(,45, 0.3)/@n(,90):a\\|_#1@n(,90)`/@n(,-90, 0.2)`\@n(,90)`/@n(,45)<`\|>\|`/`\@n(,90)<`/HO>`\|/\\|0\\/`\|`\0`\|/\`\|; #a`\|<`\>@n(,-90, 0.2)/\/\<\|>/	{}

October 16, 2026 at 4:43 am in reply to:

Calcoli di pH, 2 (SB-48)

Calcola la massa in H₃O⁺ e OH^– in …

a) … un litro di acqua. pH = 6
b) … nel lago di Constanza. Il pH è di 7

#7635

Soluzione:

Con pH = -log(c(H₃O⁺)) segue: c(H₃O⁺) = 10^-pH
Proprio così.

Con pH = -log(c(H₃O⁺)) si segue:
   c(H₃O⁺) = 10^-pH oppure c(OH^–) = 10^-pOH
   Lo stesso vale per c=n/V=m/M/V e quindi:
   m = c·V·M
a) m(H₃O⁺) = c·V·M=10^-6 mol/l · 1.0 l · 19 g/mol = 1.9·10^-5 grammi
m(OH^–) = 10^-8 mol/l · 1.0 l · 17 g/mol = 1.7·10^-7 grammi
b) Volume di Bodensee, approssimativa grossa: l= 50 km, b = 2km, h=300m
   V= 30km³=30·(1000m·1000m·1000m)=10⁹ m³ = 10¹² litri
   m(H₃O⁺) = c·V·M=10^-7 mol/l · 10¹² l · 0.019 kg/mol = 1900 kg
   m(OH^–) = 10^-7 mol/l · 10¹² l · 0.017 kg/mol = 1700 kg

October 16, 2026 at 4:43 am in reply to:

Calcoli di pH, 1 (SB-47)

Calcolare i pH delle seguenti soluzioni, se il solvente è sempre acqua pura:

1 g di acido salicylico (HCl) soluto in un litro di acqua
1 kg di acido salico (HCl) soluto in un cubo di acqua
1 g di NaOH, soluto in 2 litri di acqua
81 mg HBr in 0.1 litri di acqua
1 grammi acido solforico in 1 m³ acqua. Supposizione: completa deprotonizzazione dell'acido solforico.

#7633

Soluzione:

Nota 1: L'HCl è un acido forte (pKs < 0), cioè si dissolve completamente in acqua in H₃O⁺ e Cl^–.
Nota 2: pH = -log(c(H₃O⁺)) = -log(n/V)
Nota 3: pH + pOH = 14
Nota 4: HBr è anche un acido forte

a) pH = -log(1/36.5/1) = 1.56
b) pH = -log(1000/36.5/1000) = 1.56
c) pOH = -log(1/40/2) = 1.90, pH = 14 – pOH = 12.1
d) pH = -log(0.081/81/0.1) = 2
e) Acido solforico, H₂SO₄ emette due H⁺!
pH = -log(2·1/98/1000) = 4.69

October 16, 2026 at 4:43 am in reply to:

Konjugiert_2 (SB-11)

Indicate following acid-base chemistry-pairs more accurately ('x is conjoined acid / base of y and so on'). Specifically holds it thus, at the exercise a) or b) each all four components with each other in relationship to to set.
a) H₂CO₃ + 2 H₂O ⇄ CO2-3 + 2 H₃O⁺
b) NH₃ + H₂O ⇄ NH+4 + OH^–

#9597

Solution:

(a) H₂CO₃ is the conjugate acid to CO2-3
   CO2-3 is the conjoined base to H₂CO₃
   H₂O is the conjugate base to H₃O⁺
   H₃O⁺ is the conjugate acid to H₂O
(b) NH₃ is the conjugate base to NH+4
   NH+4 is the conjugate acid to NH₃
   H₂O is the conjoined acid to OH^–
   OH^– is the conjugate base to H₂O

October 16, 2026 at 4:43 am in reply to:

Nomenclature, déterminer les noms moléculaires (oc-12)

Les structures suivantes de différentes molécules doivent être indiquées.Appliquez les règles de nomenclature aussi précisément que possible et donnez le nom systématique.

{a) } -_(P3)_(P3)<`/>`\

{b) } /<`|><_(A75)>\/

{c) } \/<`|>\/\<_p3><\>/

{d) } @:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(white)\"&n\">@; @n(9,90)/@n(8,90)\@n(7,90)/@n(6,90)\@n(5,-90)|@n(4,90)<`/|>\@n(3,-90)<|>/@n(2,90)<\>`|@n(1,-90)

{e) } @:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(white)\"&n\">@; @n(5,90)/@n(4,90)<`|O|>\@n(3,90)/@n(2,90)\@n(1,-90)|O`|

{f) } \/<`|>\<|<`/>\NH2>/OH

{g) } /\/<`|>\|<`/>\<|OH>/\<|>/NH2

{h) } HO/<`|O|>\/<`|>\<|`/<`\H2N>|<`/>\>/\

{i) } /<`|>\/\/

{j) } /<`|>\<|>/<`|>\

{k) } /\<|>/\/\

{l) } /\<_(A150)><|OH>/\

{m) } Br\<|>/\/

#5667

Solution:

{a) } -_(P3)_(P3)<`/>`\	{1,1-Dimethyl-Cyclopropan}
{b) } /<`\|><_(A75)>\/	{2,2-Dimethylbutan}
{c) } @:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; @n(7,90)\@n(6,90)/<`\|>@n(5,90)\@n(4,90)/@n(3,90)\<_p3><@n(2,90)\@n(1,90)>/	{2,2,5-Tri-Methyl-Heptan}
{d) } @:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; @n(9,90)/@n(8,90)\@n(7,90)/@n(6,90)\@n(5,-90)\|@n(4,90)<`/\|>\@n(3,-90)<\|>/@n(2,90)<\>`\|@n(1,-90)	{4-Ethyl-2,3-Di-Methyl-Nonan}
{e) } @:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; @n(5,90)/@n(4,90)<`\|O\|>\@n(3,90)/@n(2,90)\@n(1,-90)\|O`\|	{4-Oxo-Pentan-al}
{f) } @:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; @n(6,90)\@n(5,90)/@n(4,90)<`\|>\@n(3,-90)<\|@n(2,90)<`/@n(1,90)>\NH2>/OH	{2-Amino-4-Methyl-Hexan-3-ol}
{g) } @:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; @n(10,90)/@n(9,90)\@n(8,90)/<`\|>@n(7,90)\@n(6,-90)\|@n(5,90)<`/>\@n(4,-90)<\|OH>/@n(3,90)\@n(2,-90)<\|@n(1,90)>/NH2	{2-Amino-5,7-Di-Methyl-Decan-4-ol}
{h) } @:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; HO/@n(1,90)<`\|O\|>\@n(2,90)/@n(3,90)<`\|>\@n(4,-90)<\|`/@n(6,-90)<`\H2N>\|<`/@n(8,90)>\>/\	{6-Amino-4-Ethyl-3,7,Di-Methyl-Octan-1-Carbonsäure}
{i)} /<`\|>\/\/	{2-Methylhexan }
{j) } /<`\|>\<\|>/<`\|>\	{2,3,4-Trimethylpentan }
{k) } /\<\|>/\/\	{3-Methylheptan }
{l) } /\<_(A150)><\|OH>/\	{3-Methyl-3-pentanol }
{m) } Br\<\|>/\/	{2-Brompentan }

October 16, 2026 at 4:43 am in reply to:

Concentrazioni di idroxonio ecc (SB-46)

Occorre indicare una soluzione A con pH = 7 e una soluzione B con pH = 8. Valutazioni (giusto – sbagliato) seguenti dichiarazioni:

a) Il H₃O⁺-Konzentration di A è dieci volte più grande di quello del soluzione di B
b) La H₃O⁺-Konzentration di A è pari alla concentrazione di OH^– di A
c) La H₃O⁺-Konzentration del B è dieci volte inferiore a quella del c(OH^–) del soluzione del B

#7631

Soluzione:

a) corretto
pH = 7 → c(H₃O⁺) = 10^-7 mol/l;
pH = 8 → c(H₃O⁺) = 10^-8 mol/l
b) corretto
In caso specifico, pH=7, c(H₃O⁺) = c(OH^–)
c) errato.
   pH=8 → c(H₃O⁺) = 10^-8 mol/l;
   pOH = 14-pH = 6 → c(OH^–) = 10^-6 mol/l
   Quindi c(H₃O⁺) è 100· volte più piccolo di c(OH^–)

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{a) } /\/`\|O\|\OH	⇒	/\/`\|O\|\H	⇒	/\/`\|OH	⇒	/\/
{b) } \<\|>/`\|O\|\H
{c) } \/<`\|>\\|O`\|/OH
{d) } /<`\|>\/<`\|O\|>\/
{e) } OH\|_(A75);/#2\