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- Logarithm (SB-42)Why is the logarithm used?
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in: Säure-Base_en
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- 5 days, 22 hours ago
- _siehe_auch_003 (nolink-003)[traduction manquante]
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- 5 days, 22 hours ago
- Disegnare diversi tipi di alcol (oc-04)Disegnate tre alcoli diversi:
- alcol primario
- alcol secondario
- alcol terziario
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- 5 days, 22 hours ago
- pH value 1 (SB-43)Calculate the pH of a 3-molar acid solution.
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- 5 days, 22 hours ago
- _siehe_auch_099 (nolink-099)espace réservé
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- 5 days, 22 hours ago
- Disegni di alcoli diversi, 2 (oc-05)Ogni molecola da disegnare deve avere quattro atomi di carbonio.
…. Alcol primario, alcol secondario e alcol terziario.Started by:
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- 5 days, 22 hours ago
- 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 Started by:
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- 5 days, 22 hours ago
- Nomenclatura, disegno delle molecole, 1 (oc-10)Ci sono 6 molecole, una delle quali è chiaramente diversa da tutte le altre, quale e perché?{a) } \<|>/<`|>\/\/\{b) } /<`|>\<|>/\/\|{c) } \/\/\<|<`/>\>/{d) } /<`|>\<|>/\/\/{e) } \/\|`/<`\>|<`/>\{f) } \|`/|<`/>\/<`|>\
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- 5 days, 22 hours ago
- Acid, neutral and basic, specify area (SB-45)Complete the gaps in the table with the values < 7, = 7, > 7, < 10-7, = 10-7 and > 10-7
name and address Acidic solution Neutral solution Basic solution c(H3O+ mol/l) pH c(OH– mol/l) pOH Started by:
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- 5 days, 22 hours ago
- 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
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- 5 days, 22 hours ago
- 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
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- 5 days, 22 hours ago
- 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\<|>/\/
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- 5 days, 22 hours ago
- 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.
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- 5 days, 22 hours ago
- 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`|<`\>/\/\<|>/
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- 5 days, 22 hours ago
- 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.Started by:
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- 5 days, 22 hours ago
- 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
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- 5 days, 22 hours ago
- 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\Started by:
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- 5 days, 22 hours ago
- 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
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- A 0.03 molar HCl solution is diluted a hundred times.
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- 5 days, 22 hours ago
- Strada di ossidazione 2 (oc-31)Oxidare quanto più possibile gli alcoli seguenti.{a) } /\OH{b) } \<|>/\OH{c) } \/<`|>\/OH{d) } /\<|HO>/{e) } \/<`|OH>\|{f) } /<`|>\/<`|OH>\/{g) } OH|_(A75);/#2\{h) } OH|_(A75);/#2\/`|{i) } OH|_(A75);/#2\/\|
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- 5 days, 22 hours ago
- 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 }
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- 5 days, 22 hours ago
- Strada di ossidazione 3 (oc-32)Sono state indicate le seguenti alcool. Oxidificanti – se: possibili – tutti alcolici una volta.{a) } /\OH{b) } \<|>/\OH{c) } \/<`|>\/OH{d) } HO\/<`|OH>\/OH{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/\
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- 5 days, 22 hours ago
- 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.
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- 5 days, 22 hours ago
- 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\Started by:
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- 5 days, 22 hours ago
- 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,
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- 5 days, 22 hours ago
- 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
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- 5 days, 22 hours ago
- 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.
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- 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
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- 5 days, 22 hours ago
- 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/secondStarted by:
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- 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+ H2OStarted by:
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- 5 days, 22 hours ago
- 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 kgStarted by:
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- 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)?
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- 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 nmStarted by:
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- 5 days, 22 hours ago
- 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
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- 5 days, 22 hours ago
- _siehe_auch_001 (nolink-001)[traduzione mancante]
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- 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
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- 5 days, 22 hours ago
- _siehe_auch_002 (nolink-002)[traduzione mancante]
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- 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]
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- 5 days, 22 hours ago
- _siehe_auch_003 (nolink-003)[traduzione mancante]
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- Draw different types of alcohol (oc-04)Draw three different alcohols:
- primary alcohol
- secondary alcohol
- tertiary alcohol
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- _siehe_auch_099 (nolink-099)segnaposto
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- 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.Started by:
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- 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?{a) } \<|>/<`|>\/\/\{b) } /<`|>\<|>/\/\|{c) } \/\/\<|<`/>\>/{d) } /<`|>\<|>/\/\/{e) } \/\|`/<`\>|<`/>\{f) } \|`/|<`/>\/<`|>\
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- 5 days, 22 hours ago
- 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
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- 5 days, 22 hours ago
- 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\<|>/\/
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- 5 days, 22 hours ago
- 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`|<`\>/\/\<|>/
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- 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.Started by:
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- 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\Started by:
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- 5 days, 22 hours ago
- Oxidation road 2 (oc-31)Oxidize the following alcohols as much as possible.{a) } /\OH{b) } \<|>/\OH{c) } \/<`|>\/OH{d) } /\<|HO>/{e) } \/<`|OH>\|{f) } /<`|>\/<`|OH>\/{g) } OH|_(A75);/#2\{h) } OH|_(A75);/#2\/`|{i) } OH|_(A75);/#2\/\|
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- Oxidation road 3 (oc-32)The following alcohols are prescribed. Oxidants – where applicable possible – all alcohol one time.{a) } /\OH{b) } \<|>/\OH{c) } \/<`|>\/OH{d) } HO\/<`|OH>\/OH{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/\
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