Chemieaufgaben.ch

Acceptance of 100 grams, therefore 12.79 g C, 2.15 g H and 85.07 g Bromine
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
{So what do you mean? } CH2Br { But it can’t be drawn, octet rule}
{thus} (CH2Br)2 {or} C2H4Br2
Characteristically, there are two ways to make the bromine atoms at the same C or per carbon atom per one bromateon
Br\/\Br { } /<`|Br>\Br

October 16, 2026 at 4:44 am in reply to:

Functional groups react with each other, condensation reactions 3 (oc-42)

Es be following products of condensation reaction given. Draw the skeletal shapes of the corresponding reactants analogously to the first example. Note that the notation was made with a retrosynthesis arrow. Go ahead by analogy.

{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

#9685

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 4:44 am in reply to:

Functional groups react with each other, condensation reactions 2 (oc-41)

[translation unavailable] at each exercise the following masses before:
– draw the required molecules with the functional groups
– let the two molecules drawn react with each other
– In addition to the product water, a new functional group is created. Which one?
– each partial task should at least three times dissolved are
Note: the molecules are to be neutral

Two identical alcohols react with each other
Two Different Alcohols React to Each Other
An alcohol reacts with a carboxylic acid
Two identical carbonic acids react with each other
Two different carbohydrates react with each other
An amine reacts with an carboxylic acid

#9683

Solution:

{(a) Two equal alcoholic beverages}		{ether}
{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) Two different alcohols}		{ether}
{b1) } /\OH + HO/	->	/\O/ + H2O
{b2) } /<`\|OH>\ + /\OH	->	/<`\|>\O/\ + H2O
{b3) } \<\|>/<`\|OH>\ + /\OH	->	\<\|>/<`\|>\O/\ + H2O
{(c) Alcohol + carboxylic acid}		{ester}
{c1) } \/\OH + HO/<`\|O\|>\	->	\/\O/<`\|O\|>\ + H2O
{c2) } \<\|>/<`\|OH>\ + HO/<`\|O\|>\<\|>/	->	\<\|>/<`\|>\O/<`\|O\|>\<\|>/ + H2O
{c3) } /\<\|O`\|>/OH + \<\|>/<`\|OH>	->	/\<\|O`\|>/O\/<`\|>\ + H2O
{(d) Two equal carbonic acids}		{carboxylic acid anhydride}
{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) Two different carbohydrates}		{carboxylic acid anhydride}
{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 + carboxylic acid:}		{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 4:44 am in reply to:

Functional groups react with each other, condensation reactions 1 (oc-40)

The following molecules are given. Let them react with each other in a condensation reaction.

{(a) Two equal alcoholic beverages }

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

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

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

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

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

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

{(b) Two different alcohols }

{b1) } /\OH + HO/

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

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

{(c) Alcohol and a carboxylic acid }

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

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

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

{(d) Two equal carboxylic acid }

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

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

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

{(e) Two different carboxylic acid }

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

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

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

{f) Amine and an carboxylic acid }

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

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

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

#9681

Solution:

Solutions see at exercise oc-41

October 16, 2026 at 4:44 am in reply to:

Retrosynthesis, reduction (oc-35)

The following structures are given: how can the substances be produced from the corresponding hydrocarbon plants? write down the corresponding (stable) intermediate stages. Some of these must be reduced (or oxidized) at the right place, but these technical difficulties should not be taken into account at this point.
Note also that the notation was made with a retrosynthesis arrow.

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

#9679

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 4:44 am in reply to:

Oxidation road 3 (oc-32)

The following alcohols are prescribed. Oxidants – where applicable possible – all alcohol one time.

{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/\

#9677

Solution:

{a) } /\OH	-> /`\|O\|
{b) } \<\|>/\OH	-> \<\|>/<`\|O\|>\H
{c) } \/<`\|>\/OH	-> \/<`\|>\<\|O`\|>/H {or } \/<`\|>\\|O`\|
{d) } HO\/<`\|OH>\/OH	-> O\\/<`\|O\|>\//O
{e) } `\|/:a\\|_#1`/`\`/<`\|>\|`/`\<`/HO>`\|/\\|0\\/`\|`\0`\|/\`\|; #a`\|<`\>/\/\<\|>/	-> `\|/:a\\|_#1`/`\`/<`\|>\|`/`\<`/O/>`\|/\\|0\\/`\|`\0`\|/\`\|; #a`\|<`\>/\/\<\|>/
{f) } `/`-`\/-<\>\-OH {Oxidation not possible, no 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 4:44 am in reply to:

Oxidation road 2 (oc-31)

Oxidize the following alcohols as much as possible.

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

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

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

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

#9675

Solution:

{a) } /\OH	->	/\O`\	->	/<\O`\>`\|OH
{b) } \<\|>/\OH	->	\<\|>/\O`\	->	\<\|>/<\O`\>`\|OH
{c) } \/<`\|>\/OH ->	->	\/<`\|>\/O`/	->	\/<`\|>\<\|OH>
{d) } /\<\|HO>/	->	/\<\|O`\|>/	->	{No further oxidation is possible}
{e) } \/<`\|OH>\\|	->	\/<`\|O\|>\\|	->	{No further oxidation is possible}
{f)} /<`\|>\/<`\|OH>\/	->	/<`\|>\/<`\|O\|>\/	->	{No further oxidation is possible}
{g) } OH\|_(A75);/#2\	->	{No oxidation is possible}
{h) } OH\|_(A75);/#2\/`\|	->	{No oxidation is possible}
{i) } OH\|_(A75);/#2\/\\|	->	{No oxidation is possible}

October 16, 2026 at 4:44 am in reply to:

Oxidation road 1 (oc-30)

Oxidize the following substances once. At each oxidation, give the only short-term Intermediate product. Example:

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

#9673

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\				{Oxidation not possible}

! Attention ! d) contains another error, bug?! in charchem?!

October 16, 2026 at 4:44 am in reply to:

Nomenclature, Chirality, 2 (oc-21)

Draw three molecules each with at least one chiral center, the molecules drawn must be neutral, the chiral centers each marked with a star.
(a) all elements are permitted
(b) elements C, H and Br must occur at least once in each molecule; other elements are not permitted.

#9671

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 4:44 am in reply to:

Nomenclature, Chirality (oc-20)

If the molecule contains one or more chiral centres, mark them with an asterisk; if no chiral centres are to be found, indicate kC (no chiral centres).

{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`|<`\>/\/\<|>/

#9669

Solution:

{a: } /<`\|>\/\/	{[translation unavailable]}
{b: } /<`\|>\<\|>/<`\|>\	{c c c}
{c: } @:n(n,a,L:.2)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; /\@n(*,-90)<\|>/\/\	{}
{d: } /\<_(A150)><\|OH>/\	{[translation unavailable]}
{e: } @:n(n,a,L:.2)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; Br\@n(*,-90)<\|>/\/	{}
{f: } OH`\`\|O\|`/\|`/`\`\|//\\|\O\|`/O/\	{[translation unavailable]}
{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 4:44 am in reply to:

Nomenclature, determining molecular names (oc-12)

The following structures of different molecules are given, apply the nomenclature rules as accurately as possible and give the systematic name.

{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\<|>/\/

#9667

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 4:44 am in reply to:

Nomenclature, molecules draw, 2 (oc-11)

The following names of various molecules are given. Draw the corresponding molecule (spelling).

hexane
2-methyl octane
3,4-[translation unavailable]-Ethyl-2,2,5-tri-methyl decane
Ethanol (also Ethan-1 -ol)
Propan-1,2,3-[translation unavailable]
Butan-1-other
Butan-2-[translation unavailable]
[translation unavailable]
4-[translation unavailable]-Pentan-1-[translation unavailable]
2-[translation unavailable]-Pentan-2-[translation unavailable]
3-[translation unavailable]-Heptan-1-[translation unavailable]
3-hydroxy butaneal
2,3-[translation unavailable]-Butan-2,3-diol
2,3-dimethyl pentane
1 Ethyl-1-methyl cyclobutan
3-Ethyl-2,2,4-[translation unavailable]-hexane

#9665

Solution:

{a) hexane }	/\/\/
{(b) 2-methyl octane: }	@:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; @n(8,90)/@n(7,90)\@n(6,90)/@n(5,90)\@n(4,90)/@n(3,90)\@n(2,-90)<\|>/@n(1,90)
{c) 3,4-diethyl- 2,2,5-tri-methyl decane: }	/<_q4><_p4>/<`\|/>\<\|\>/<`\|>\/\/\
{(d) Ethanol: }	/\OH
{(e) Propan-1,2,3-tri-ol:}	HO/\<\|OH>/\OH
{f) Butan-1-other: }	/\/`\|\|O
{g) Butan-2-[translation unavailable]:}	@:n(n,a,L:.4)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; @n(4,90)/@n(3,90)\@n(2,-90)<\|\|O>/@n(1,90)
{(h) Cyclohexanol}	`/`-`\/-\-OH
{i) 4-[translation unavailable]-Pentan-1-[translation unavailable]}	\<\|>/\/\OH { } { or: } HO/\/\<\|>/
{j) 2-[translation unavailable]-Pentan-2-[translation unavailable]}	HO/<_q4><_p4>/\/
{k) 3-[translation unavailable]-Heptan-1-[translation unavailable]}	HO\/\<\|NH2>/\/\
{(l) 3-hydroxy butaneal}	\<\|OH>/\//O
{(m) 2,3-dimethyl butane-2,3-diol }	HO-<_q4><_p4>–<_q4><_p4>-OH
{n) 2,3-dimethyl pentane }	/<`\|>\<\|>/\
{o) 1-Ethyl-1-methyl cyclobutan }	(`/)(`\/)\|-`\|_#1
{(p) 3 – Ethyl-2,2, 4 – Tri-methyl- hexane }	@: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 4:44 am in reply to:

Nomenclature, drawing of molecules 1 (oc-10)

The following 6 molecules are given. One of them is clearly different from all the others. What and why?

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

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

#9663

Solution:

The last molecule (f) is different from all others. (a) to (e) are identical, even if they look different. Consider the longest possible carbon chain and see which carbon atoms can be observed branching off with one carbon atom each.
The systematic nomenclature clearly helps to highlight the differences. Keep in mind that the numbering of carbon atoms does not necessarily have to be done from left to right, but in such a way that the sum of the branches is as small as possible.

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

October 16, 2026 at 4:44 am in reply to:

Draw different types of alcohol, 2 (oc-05)

Each molecule to be drawn must have four carbon atoms.
…. primary alcohol, secondary alcohol and tertiary alcohol.

#9661

Solution:

{primary alcohol: } /\/\OH

{secondary alcohol: } /<`|OH>\/

{tertiary alcohol: } –<|><`|>-OH

October 16, 2026 at 4:44 am in reply to:

_siehe_auch_099 (nolink-099)

segnaposto

#7695

Soluzione:

segnaposto

October 16, 2026 at 4:44 am in reply to:

Draw different types of alcohol (oc-04)

Draw three different alcohols:

primary alcohol
secondary alcohol
tertiary alcohol

#9659

Solution:

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

October 16, 2026 at 4:44 am in reply to:

_siehe_auch_003 (nolink-003)

[traduzione mancante]

#7693

Soluzione:

[traduzione mancante]

October 16, 2026 at 4:44 am in reply to:

Draw different functional groups, 3 (oc-03)

Draw any molecule that contains all functional groups at the same time. The basic structure should always be an alkali array.

Alkene, alcohol and carboxylic acid
Alkenes, aldehydes, ketones and amides
Carboxylic acid, alcohol and ketone
Amine, amide, carboxylic acid and [translation unavailable]

#9657

Solution:

/<`||>\<|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 4:44 am in reply to:

_siehe_auch_002 (nolink-002)

[traduzione mancante]

#7691

Soluzione:

[traduzione mancante]

October 16, 2026 at 4:44 am in reply to:

Draw different functional groups, 2 (oc-02)

Draw the structural formula for each exercise. In addition to the functional group, no other functional groups may be drawn; only carbon-hydrogen links are permitted.

A linear, not of branches alcohol with [translation unavailable] 3 carbon atoms
A branched alcohol with exactly 5 carbon atoms
All alcohol (branched, not branched) with [translation unavailable] 4 carbon atoms
Three different ethers with a total of 4 carbon atoms each
The smallest possible (as little C as possible) ketone
The smallest possible (as little C as possible) aldehyde
The smallest possible (as little C as possible) amine
The smallest possible (as little C as possible) ester

#9655

Solution:

{a) }		/\/OH
{b) }		/\<\|>/\OH
{c) }		/\/\OH { } /<`\|>\/OH { } OH\|_(A75);/#2\
{d) }		/O\/\ { } /\O/\ { } -_p_pO_p_p
{(e) The smallest ketone }		/<`\|\|O>\
{(f) Minimum aldehyde}		H/<`\|\|O>\H { or: } <`\|\|O>
{(g) Minimum amount of amine}		H2N/
{(h) Smallest ester}		<`\|O\|>\O/

October 16, 2026 at 4:44 am in reply to:

Draw different functional groups (oc-01)

Draw for each exercise each three different molecules, which consist only of one hydrocarbon skeleton (alkane skeleton, i.e. single bonds) and the following functional group:

Not branched alkanes
Branched alkanes
Cyclic alkanes
Alkenes
Alkyne
Alcohol
aldehyde
ketone
Amina
carboxylic acid
amino acid
ester
amide

#9653

Solution:

Not branched alkanes
\/\ ; { } ; \/\|\ ; { } ; \/\/\/\/
Branched alkanes
/<`|>\ ; { } ; /<`|>\/ ; { } ; \<|>/\<|>/
Cyclic alkanes
-_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
Alkenes
/=/ ; { } ; /\/\// ; { } ; \//\<|>//\
Alkyne
///c/\ ; { } ; -%-/ ; { } ; \\\c\<|>/\/
Alcohol:
\/\|OH ; { } ; \<|>/OH ; { } ; /\OH
[translation unavailable]:
\||O ; { } ; O\\|\|`//`\`||/ ; { } ; H\/O`/|\|`/`\`|<_(a-30)><`->/
Ketones:
/<`||O>\ ; { } ; /\/<`||O>\ ; { } ; /\<||O>/
[translation unavailable]:
/\NH2 ; { } ; /<`|NH2>\/ ; { } ; /\/\/\/NH2
Carbonic acids:
OH`/|O`| ; { } ; Cl`/|`/N`\\<`/Cl>`|//\|\|O`|/OH ; { } ; HO/`|O|\/\/\|O`|/OH
Amino acids:
H2N-CH<|$color(red)CH3>-$color()COOH ; { } ; |`//`\`||/\/`|O|\N<_(y.5)H>/\|O`|/OH ; { } ; \<|>/`|<`\dH2N>/`|O|\OH
ester:
|`//`\`||/\/`|O|\O/\||\//`|`\`/ { } H|`/O/\O/\|`/O|`/O/\H { } /O\|O`|/\/`|O|\O/
Amide , please .
H2N/\O { } /\|O`|/N<_(y-.5)H>\ { } -_qNH_q<`|O|>_q_q

October 16, 2026 at 4:44 am in reply to:

_siehe_auch_001 (nolink-001)

[traduzione mancante]

#7689

Soluzione:

[traduzione mancante]

October 16, 2026 at 4:44 am in reply to:

Calculation QM 3 (qm-03)

Estimate (computing) the number of photons emitted by a bulb with 100 W per second in the visible range of light.
Note 1: Simplify that λ = 500 nm

#9651

Solution:

E=h·f = h·c/λ
E_total = x·h·f = x·h·c/λ
   E_total equal to 100 W/s
   x is the number of particles at 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, i.e. about 1/1000 mol

October 16, 2026 at 4:44 am in reply to:

Struttura previsioni, esercizio 1 (oc-50)

È stato dato il seguente risultato di un analisi elementare: C 12.79%. H 2.15%, il resto è bromo

Che cos'è la formula molecolare
Disegnare la molecola in base alla formula molecolare. Cosa si nota?
Qual è l'aspetto di NMR delle strutture (ambedue)?

#7687

Soluzione:

Ricordiamo le percentuali: queste sono sempre percentuali di massa. Per semplificare il calcolo, facciamo 100 grammi di sostanza

Assunzione di 100 grammi: 12.79 g di C, 2.15 g di H e 85.07 g di Bromo
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
{Quindi… } CH2Br { Lascia si ma non disegnare, regola dell’ottetto}
{quindi} (CH2Br)2 {oppure} C2H4Br2
In grafico, ci sono due possibilità per i bromo atomi allo stesso C o per ogni atomo di carbonio per un bromatone
Br\/\Br { } /<`|Br>\Br

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

Calculation QM 2 (qm-02)

Electrons of the solar wind reach Earth after a flight time of about 30 hours, where they can produce polar lights.
In the case of polar lamps, green illumination is predominant, caused by the stimulation of oxygen at λ = 557.7 nm.
Calculate that the kinetic energy of the electrons of the solar wind is sufficient to stimulate oxygen.
Note: The distance between Earth and the Sun is 150 million km
Note 2: m(e^–)= 9.1·10^-31 kg

#9649

Solution:

· General idea: the kinetic energy must correspond to at least excitation energy
· 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 = approx. 2.4
i.e. there is more kinetic energy than excitation energy

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

Gruppi funzionali, reazioni di condensazione 3 (oc-42)

Es siano seguenti prodotti di reazione di condensazione dato. Tracciate analogamente, per esempio, le forme scheletriche degli reagenti. Si noti che la notazione è stata fatta con una freccia di retrosintesi. Andate analogamente.

{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

#7685

Soluzione:

{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:

Calculation QM 1 (qm-01)

(a) Calculate the energy of a blue light photon with a wavelength of 450 nm.
(b) Calculate the number of light particles (photons) of a blue lamp with a power of 300 watts
Watt is Joule/second

#9647

Solution:

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:

Gruppi funzionali, reazioni di condensazione 2 (oc-41)

Per ogni esercizio massa:
– disegnare le molecole richieste con i gruppi funzionali
– che le due molecole tracciate reagiscano tra loro
– Oltre al prodotto, l'acqua crea un nuovo gruppo funzionale.
– Ogni compito parziale deve essere risolto almeno tre volte
Nota: le molecole devono essere neutre

Due alcoli identici reagiscono tra loro
Due alcoli diversi reagiscono tra loro
Un alcol reagisce con un acido carbossilico
Due acidi di carbonio identici reagiscono tra loro
Due diversi acidi di carbonio reagiscono tra loro
Un amino reagisce con un acido carbossilico

#7683

Soluzione:

{a) Due alcoli uguali}		{etere}
{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) Due alcoli diversi}		{etere}
{b1) } /\OH + HO/	->	/\O/ + H2O
{b2) } /<`\|OH>\ + /\OH	->	/<`\|>\O/\ + H2O
{b3) } \<\|>/<`\|OH>\ + /\OH	->	\<\|>/<`\|>\O/\ + H2O
{c) alcol + acido carbossilico}		{estere}
{c1) } \/\OH + HO/<`\|O\|>\	->	\/\O/<`\|O\|>\ + H2O
{c2) } \<\|>/<`\|OH>\ + HO/<`\|O\|>\<\|>/	->	\<\|>/<`\|>\O/<`\|O\|>\<\|>/ + H2O
{c3) } /\<\|O`\|>/OH + \<\|>/<`\|OH>	->	/\<\|O`\|>/O\/<`\|>\ + H2O
{d) Due acidi di carbonio uguali}		{anidride carbossilica}
{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) Due diversi acidi di carbonio}		{anidride carbossilica}
{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) ammina + acido carbossilico:}		{ammide}
{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:

Gruppi funzionali, reazioni di condensazione 1 (oc-40)

Le molecole seguenti sono indicate. Lasciate che reagiscano in una reazione di condensazione.

{a) Due alcoli uguali }

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

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

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

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

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

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

{b) Due alcoli diversi }

{b1) } /\OH + HO/

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

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

{c) Alcol e una acido carbossilico }

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

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

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

{d) Due uguali acido carbossilico }

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

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

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

{e) Due differenti acido carbossilico }

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

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

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

{f) Amine e una acido carbossilico }

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

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

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

#7681

Soluzione:

Soluzioni vedi a esercizio oc-41

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

Calculation titration 1 (SB-70)

Titration of an acid (20 ml, unknown concentration) with a base (0.1 M) is required up to the equivalence point of 30 ml of the base. Calculate the concentrations of the acid.

#9645

Solution:

It holds at the same time, equivalence point: n(Base) = n(Säure)
with c=n/V we get:
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:

pH calculations, 4 (SB-62)

One liter of two solutions A and B with the same starting pH of 4.8 shall be given. A is a hydrochloric acid solution, B is an acidic acid solution.
Now one liter of water is added to each solution.

Calculate the new pH of solution A and B
In the case of a solution that has a greater change in pH,

Note: HCl is a strong acid (pKs < 0), acetic acid a weak acid (pKs > 0)

#9643

Solution:

Based on the pH value, HCl for the strong acid and acidic acid for the weak acid can now be the concentration is calculated:

strong acid: HCl
pH = -log(c(H₃O⁺)), c(H₃O⁺) = 10^-pH = 0.000015 mol/l
C = n/V or n = c · V
n = 0.000015 · 1.0 l = 0.000015 mol
Weak acid: acetic acid
pH = 1/2(pKs-log(HA)), c(HA) = …
n = c·V =
Now 1 litre of pure water is added to each solution, and with c=n/V we get the following:
HCl: pH = -log(c(H₃O⁺)) = -log(n/V)
pH = [translation unavailable]

… not finished yet !

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

Retrosintesi, riduzione (oc-35)

Le seguenti strutture sono indicate. Come si possono produrre le sostanze da tali impianti di idrogeno di carbonio? Scrivi gli intermedi (stabili) corrispondenti. In parte è necessario ridurre (o ossidare) al posto giusto, ma queste difficoltà tecniche non devono essere prese in considerazione.
Nota anche che la notazione è stata fatta con una freccia di retrosintesi.

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

#7679

Soluzione:

{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:

Strada di ossidazione 3 (oc-32)

Sono state indicate le seguenti alcool. Oxidificanti – se: possibili – tutti alcolici una volta.

{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/\

#7677

Soluzione:

{a) } /\OH	-> /`\|O\|
{b) } \<\|>/\OH	-> \<\|>/<`\|O\|>\H
{c) } \/<`\|>\/OH	-> \/<`\|>\<\|O`\|>/H {oppure } \/<`\|>\\|O`\|
{d) } HO\/<`\|OH>\/OH	-> O\\/<`\|O\|>\//O
{e) } `\|/:a\\|_#1`/`\`/<`\|>\|`/`\<`/HO>`\|/\\|0\\/`\|`\0`\|/\`\|; #a`\|<`\>/\/\<\|>/	-> `\|/:a\\|_#1`/`\`/<`\|>\|`/`\<`/O/>`\|/\\|0\\/`\|`\0`\|/\`\|; #a`\|<`\>/\/\<\|>/
{f) } `/`-`\/-<\>\-OH {Non è possibile l’ossidazione, nessun 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:

Calculations of pH, mixtures (SB-61)

Calculate the pH of the following mixtures.

For 80 ml of a 0.05 M solution of hydrochloric acid, 100 ml of a 0.01 M solution of sodium nitrate is given.
For 60 ml of a 0.015 KOH solution, 30 ml of 0.2 M saline solution is given.
10 ml 0.2 M saltic acid solution is first diluted tenfold and then replaced with 1.0 g Ca(OH)₂.

#9641

Solution:

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
   · Remaining staying at the neutralization reaction 0.003 mol HCl
   · Note: total volume is 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
   · Note that KOH behaves similarly to NaOH, so it's also a base.
   · Remaining staying at the neutralization reaction 0.0051 mol HCl
   · Please note: total volume 60ml + 30 ml
   · pH=-log(0.0051/0.09) = 1.25
· Start pH: pH=-log(0.2) = 0.7
   · Tenfold dilution I mean,, that to the 10 ml still additional 90 ml this is the. Water is given
   · 10 ml 0.2 M HCl → n(HCl)=0.002 mol or 0.002 mol
   · n(HCl)=0.002 mol → 0.002 mol H⁺
   · Total volume: 10ml+90ml=100 ml
   · pH=-log(0.002/0.1) = 1.7
   · Increase by one unit of pH makes sense at ten times dilution

   · 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
   · Neutralization reaction: 0.002 mol H and 0.027 mol OH^–
   · remaining 0.025 mol OH^–
   · Total volume: 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:

Strada di ossidazione 2 (oc-31)

Oxidare quanto più possibile gli alcoli seguenti.

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

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

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

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

#7675

Soluzione:

{a) } /\OH	->	/\O`\	->	/<\O`\>`\|OH
{b) } \<\|>/\OH	->	\<\|>/\O`\	->	\<\|>/<\O`\>`\|OH
{c) } \/<`\|>\/OH ->	->	\/<`\|>\/O`/	->	\/<`\|>\<\|OH>
{d) } /\<\|HO>/	->	/\<\|O`\|>/	->	{non è possibile ulteriore ossidazione}
{e) } \/<`\|OH>\\|	->	\/<`\|O\|>\\|	->	{non è possibile ulteriore ossidazione}
{f)} /<`\|>\/<`\|OH>\/	->	/<`\|>\/<`\|O\|>\/	->	{non è possibile ulteriore ossidazione}
{g) } OH\|_(A75);/#2\	->	{non è possibile l’ossidazione}
{h) } OH\|_(A75);/#2\/`\|	->	{non è possibile l’ossidazione}
{i) } OH\|_(A75);/#2\/\\|	->	{non è possibile l’ossidazione}

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

Neutralizations (SB-60)

Form the neutralization reactions in ion type and form the corresponding salts, analogous to exercise a).

{(a) Sodium hydroxide + hydrochloric acid }	Na^+ + OH^- + H3O^+ + Cl^- -> NaCl{([translation unavailable])} + 2H2O
{(b) Sodium hydroxide + hydrochloric acid }
{(c) Sodium hydroxide + carbonic acid }
{(d) Potassium hydroxide + sulphuric acid }
{(e) Potassium hydroxide + phosphoric acid }

#9639

Solution:

{(a) Sodium hydroxide + hydrochloric acid }	Na^+ + OH^- + H3O^+ + Cl^- -> NaCl{([translation unavailable])} + 2H2O
{(b) Sodium hydroxide + hydrochloric acid }	Na^+ + OH^- + H3O^+ + NO3^- -> NaNO3{([translation unavailable])} + 2H2O
{(c) Sodium hydroxide + carbonic acid }	2Na^+ + 2OH^- + 2H3O^+ + 2CO3^2- -> Na2CO3{([translation unavailable])} + 4H2O
{(d) Potassium hydroxide + sulphuric acid }	2K^+ + 2OH^- + 2H3O^+ + SO4^2- -> K2SO4{([translation unavailable])} + 4H2O
{(e) Potassium hydroxide + phosphoric acid }	3K^+ + 3OH^- + 3H3O^+ + PO4^3- -> K3PO4{([translation unavailable])} + 4H2O

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

Calculations of pH, dilutions (SB-49)

Calculate the pH of the following solutions if the solvent is always pure water:

A 0.03 molar HCl solution is diluted a hundred times.
Calculate the pH before and after dilution
20 ml 0.05 M HCl is diluted with 100 ml distilled water.
Calculate the pH before and after dilution
50 ml of 0.2 mo of NaOH solution is given 100 ml of distilled water.
Calculate the pH before and after dilution

#9637

Solution:

Start- pH: pH=-log(0.03) = 1.5
   A hundred times dilution means that 99 parts of a solution are water.
   z.B. 1 ml the 0.03 M solution [translation unavailable] 99 ml water
   · 1000 ml ≙ 0.03 mol H⁺
   · 1 ml → 0.00003 mol H⁺
   · pH=-log(0.00003/0.001)=1.52
   · Note that 99 ml of water is allowed
   · pH=-log(0.00003/(0.001+0.099)) = 3.52
   It should be noted that the increase of two decimal places in
   H⁺ Concentration changes whether by 2 units at pH
Start- pH: pH=-log(0.05) = 1.3
   · 20 ml 0.05 M HCl is called:
   · 1000 ml ≙ 0.05 mol H⁺
   · 20 ml → 0.001 mol H⁺
   · Note that 100 ml of water is allowed
   · the new pH: pH=log(n/V)=-log(0.001/(0.02+0.1))= 2.079
Start pH: pOH=-log(0.2)=0.7, pH=14-0.7=13.3
   · 1000 ml ≙ 0.2 mol OH^–
   · 50 ml → 0.01 mol OH^–
   · Note that 100 ml of water is allowed
   · 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:

Strada di ossidazione 1 (oc-30)

Oxidare le seguenti sostanze una volta ogni volta. Per ogni ossidazione si deve indicare l’ossidazione a breve termine. Prodotto intermedio: esempio:

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

#7673

Soluzione:

{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\				{Non è possibile l’ossidazione}

! D) contiene un altro errore, bug?! in charchem?!

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

pH calculations, 2 (SB-48)

Calculate the mass at H₃O⁺ and OH^– in …

a) … one litre of water. pH = 6
(b)… in Lake Constance. The pH was 7

#9635

Solution:

With pH = -log(c(H₃O⁺)) following:: c(H₃O⁺) = 10^-pH
It is the same.

The pH = -log(c(H₃O⁺)) is as follows:
   c(H₃O⁺) = 10^-pH or c(OH^–) = 10^-pOH
   Similarly, c=n/V=m/M/V and thus:
   m = c·V·M
(a) m(H₃O⁺) = c·V·M=10^-6 mol/l · 1.0 l · 19 g/mol = 1.9·10^-5 grams
m(OH^–) = 10^-8 mol/l · 1.0 l · 17 g/mol = 1.7·10^-7 grams
b) volumes lake Constance, coarse estimated: l= 50 km, b= 2km, h=300m
   V= 30km³=30·(1000m·1000m·1000m)=10⁹ m³ = 10¹² litres
   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:

Nomenclatura, chiralità, 2 (oc-21)

Disegnare tre molecole che abbiano almeno un centro di chiralità. Le molecole disegnate devono essere neutre. Indicare i centri di chiralità con una stella.
a) tutti gli elementi sono ammessi
In ogni molecola, gli elementi C, H e Br devono essere presenti almeno una volta, altri elementi non sono ammessi.

#7671

Soluzione:

{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:

Nomenclatura, chiralità (oc-20)

Sono state indicate le seguenti molecole: Se la molecola contiene uno o più centri di chiralità, segnale con una stella. Se non si trovano centri di chiralgia, indicare kC (nessun centro di chiralgia).

{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`|<`\>/\/\<|>/

#7669

Soluzione:

{a: } /<`\|>\/\/	{[traduzione mancante]}
{b: } /<`\|>\<\|>/<`\|>\	{[traduzione mancante]}
{c: } @:n(n,a,L:.2)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; /\@n(*,-90)<\|>/\/\	{}
{d: } /\<_(A150)><\|OH>/\	{[traduzione mancante]}
{e: } @:n(n,a,L:.2)<_(A&a,L&L,N0)$itemColor1(blue)\"&n\">@; Br\@n(*,-90)<\|>/\/	{}
{f: } OH`\`\|O\|`/\|`/`\`\|//\\|\O\|`/O/\	{[traduzione mancante]}
{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:

pH calculations, 1 (SB-47)

Calculate the pH of the following solutions if the solvent is always pure water:

1 g hydrochloric acid (HCl) dissolved in a litres water
1 kg of hydrochloric acid (HCl) dissolved in one cubic meter of water
1 g NaOH, dissolved in 2 litres of water
81 mg HBr in 0.1 litres of water
1 grams sulphuric acid in 1 m³ water. Assumption: complete deprotonation of the sulphuric acid.

#9633

Solution:

Note 1: HCl is a strong acid (pKs < 0), i.e. it completely decomposes in water to H₃O⁺ and Cl^–.
Note 2: pH = -log(c(H₃O⁺)) = -log(n/V)
Note 3: pH + pOH = 14
Note 4: HBr is also a strong acid

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) Sulphuric acid, H₂SO₄ gives off two H⁺!
pH = -log(2·1/98/1000) = 4.69

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

Nomenclatura, determinazione dei nomi molecolari (oc-12)

Indicare le seguenti strutture di varie molecole, applicare le regole di nomenclatura con la massima precisione possibile e indicare il nome sistematico.

{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\<|>/\/

#7667

Soluzione:

{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:

Concentrations of hydroxonium etc (SB-46)

Given be a solution A with pH = 7 and a solution B with pH = 8. Assessment (correct – wrong) of the following statements:

(a) The H₃O⁺-Konzentration of A is ten times larger than that of solution of B
(b) The H₃O⁺-Konzentration of A is equal to the OH^– concentration of A
(c) The H₃O⁺-Konzentration of B is ten times smaller than the c(OH^–) of solution of B

#9631

Solution:

(a) correctly
pH = 7 → c(H₃O⁺) = 10^-7 mol/l;
pH = 8 → c(H₃O⁺) = 10^-8 mol/l
(b) correct
Specifically at pH=7, c(H₃O⁺) = c(OH^–)
(c) wrong.
   pH=8 → c(H₃O⁺) = 10^-8 mol/l;
   pOH = 14-pH = 6 → c(OH^–) = 10^-6 mol/l
   So c(H₃O⁺) is 100· times smaller than c(OH^–)

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)

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

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