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- Calculer différentes valeurs (pH, etc) (SB-44)Calculer / compléter les valeurs dans le tableau suivant.
c(H3O+ mol/l) c(OH– mol/l) pH pOH 0.001 0.01 0.1 0.2 12 11 Started by:
in: Säure-Base_fr
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- Comportamento di un acido o di una base (SB-03)Formulare le reazioni per le seguenti sostanze indicate. L'abbreviazione LM è per solvente.
Non si tratta di valutare se Le richiesto reazioni possibili sono o non, ma per questo motivo, che z.B. A una richiesto reazione acida il dato particolato un H+ trasmette e il soluzioni questo e poi accoglie.- a) HBr, reazione acida, LM: acqua
- b) HF, reazione di base, LM: acqua
- (c) HF, reazione acida, LM: acqua
- d) HF, reazione acida, LM: NH3
- e) H2O, reazione acida, LM: acqua
- f) CH3COOH, reazione acida, LM: acqua
- g) CH3OH, reazione di base, LM: NH3
- h) CH3OH, reazione acida, LM: acqua
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- Comparing entropy 2 (thermo-12)Given be the molecule C2H6 and CH4. Which this both connections is shown a higher entropy on. Reason your choice.
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- Acide, neutre et de base, indiquer la zone (SB-45)Compléter les lacunes dans le tableau avec les valeurs < 7, = 7, > 7, < 10-7, = 10-7 et > 10-7
nom Solution acide solution neutre Solution de base c(H3O+ mol/l) pH c(OH– mol/l) pOH Started by:
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- Konjugiert_1 (SB-10)Che cosa sono le coppie acido-base coniugate?
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- Calculation of reaction enthalpy 1 (thermo-13)Calculate the standardized-reaction enthalpy following reaction:
H2(g) + 1/8 S8 (s) ⇄ H2S(g)Started by:
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- Concentrations d’hydroxonium, etc (SB-46)Une solution A avec pH = 7 et une solution B avec pH = 8 doivent être fournies. Les évaluations (c'est vrai – fausse) suivantes déclarations:
- a) les H3O+-Konzentration de A est dix fois ainsi grosse comment les le solution B
- b) les H3O+-Konzentration de A est égal le OH– concentration de A
- c) les H3O+-Konzentration de B est dix fois plus petit comme les c(OH–) le solution B
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- Konjugiert_2 (SB-11)Indicare seguenti acidi-basi-coppie più precisamente ('x è coniugi acido / base di y [traduzione mancante]'). Concretamente vale esso quindi, a il esercizio a) oppure b) ciascuno tutti quattro componenti con gli altri in relazione a sette.
a) H2CO3 + 2 H2O ⇄ CO2-3 + 2 H3O+
b) NH3 + H2O ⇄ NH+4 + OH–Started by:
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- Calculation of reaction enthalpy 2 (thermo-14)At the turn of 60 grams of grape sugar, x kJ is released. Calculate the ΔHof from this.
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- Calculs du pH, 1 (SB-47)Calculer les pH des solutions suivantes, le solvant étant toujours de l'eau pure:
- 1 g d'acide salé (HCl) dissous dans un litre d'eau
- 1 kg d'acide salé (HCl) dissous dans un mètre cube d'eau
- 1 g de NaOH, dissous dans 2 litres d'eau
- 81 mg HBr dans 0.1 litre d'eau
- 1 grammes l'acide sulphurique dans 1 m3 eau. Supposons une déprotonation complète de l'acide sulfurique.
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- Conjugato, diversi esempi 1 (SB-12)Completare la tabella seguente, sollevando il primo esempio.
Dovrebbe esso per una substanza z.B. Nessuno coniugi base dare (z.B. F-), e poi dovrebbe il contenuto delle cellule deve essere dotato di un 'X'.conj. base coni. acido H2O OH– H3O+ HS– F– H2PO–4 PO-34 H3PO4 NH3 NO–2 Started by:
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- Calculation of reaction enthalpy 3 (thermo-15)Glucose (C6H12O6) can be implemented in different ways: aerobic (with O2) and anaerobic (without O2).
For both reactions, formulate the reaction equation and calculate the standard reaction enthalpy.
Note I: if the ΔHof is unknown, then generally calculate x, y, etc.
Note II: state of matter stateStarted by:
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- Calculs du pH, 2 (SB-48)Calculer la masse à H3O+ et OH– en …
- a) … un litre d'eau. pH = 6
- b) … dans le lac de Constance. Le pH est de 7
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- Conjugato, diversi esempi 2 (SB-13)Indicare le seguenti reazioni. Ordina le sostanze coinvolte in parti corrispondenti acido – basi{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^-
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- Calculation of reaction enthalpy 4 (thermo-16)A cold bag is made up of the following ingredients: 25 g NH4NO3 (ammonium nitrate) and 75 ml of water. Now the ammonium nitrate is dissolved in water.
Note 1: the (specific) heat capacity cw of the water is 75.3_nbspJ/(K·mol) or 4.18 J/(K·kg).
Note 2: Q = m·c·ΔT
Note 3: The ammonium nitrate dissolves in water in its ions, the name indicating which.Started by:
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- pH-calculs, diluctions (SB-49)Calculer les pH des solutions suivantes, le solvant étant toujours de l'eau pure:
- Une 0.03 moulés HCl-solution est cent fois dilué.
Calculer le pH avant et après dilution - 20 ml 0.05 M HCl est dilué avec 100 ml d'eau distillée.
Calculer le pH avant et après dilution - À 50 ml une 0.2 molars NaOH-solution est 100 ml destillés eau donné.
Calculer le pH avant et après dilution
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- Une 0.03 moulés HCl-solution est cent fois dilué.
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- Acidi nell’acqua (SB-14)I seguenti acidi sono dissoluti in acqua. Es fabbricazione le accettazione, che le acidi completamente deprotoniare. Scrivere la corrispondente ioni analogamente a esercizio (a).{a) Acido acido: }CH3COOH + H2O <=> CH3COO^- + H3O^+{b) Acido salino:}{c) Acido formico: }{(d) acido solforico: }{(e) Acido salpetrico: }{(e) Acido di carbonio: }
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- Calculation of reaction enthalpy 5 (thermo-17)To sweat and cut, you need fuel gas. You usually add ethane (C2H2). Why not the cheaper methane (CH4)?
Note: For both gases, formulate the combustion reaction and determine the reaction enthalpy per (!) molecule C2H2 and CH4 respectively.
ΔHof(CH4) = -75 kJ/mol
ΔHof(C2H2) = 227 kJ/mol
ΔHof(H2O) = -285 kJ/mol
ΔHof(CO2) = -393 kJ/molStarted by:
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- Neutralisations (SB-60)Le formulaire les réactions de neutralisation dans type d’écriture et des images les d’autres sels, par exemple: à l’intérieur exercice a).{a) Hydroxyde de sodium + acide salé }Na^+ + OH^- + H3O^+ + Cl^- -> NaCl{(l’accès à l’eau)} + 2H2O{b) Hydroxyde de sodium + acide salphétrique }{(c) Hydroxyde de sodium + acide carbonique }{d) Hydroxyde de potassium + acide sulphurique }{e) Hydroxyde de potassium + acide phosphorique }
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- Disegnare diversi acidi (SB-15)Disegna le formule di Lewis dei seguenti acidi.{a) Acido acido: }{b) Acido salino:}{c) Acido formico: }{(d) acido solforico: }{(e) Acido salpetrico: }{(e) Acido di carbonio: }
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- pKs (SB-20)Che cos'è il pKs?
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- Calculation of reaction enthalpy 6 (thermo-18)Calculate the ΔHof of carbon monoxide and consider the following reaction:
CH4(g) + H2O(l) → CO(g) + 3 H2(g). The (experimental) ΔHoR for this purpose of which: 249.5 kJ/mol.Started by:
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- pH-calculs, des mélanges (SB-61)Calculez le pH des mélanges suivants.
- Pour 80 ml d'une solution d'acide salé de 0.05 M, 100 ml d'une solution de sodium de 0.01 M doivent être administrés.
- Pour 60 ml de solution de 0.015 KOH, 30 ml de solution de 0.2 M d'acide salé est administré.
- 10 ml 0.2 M solution de l'acide salé sont tout d'abord dix fois dilué et alors avec
1.0 g Ca(OH)2 déposé.
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- pH – pKs (SB-21)Il pH e il pKs sono uguali?
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- Calculation of reaction enthalpy 7 (thermo-19)Adopted, one decides itself for the not health variation, that the complete daily energy intake by means of drinking of alcohol (40% alcohol) implemented is. The following assumptions and specifications shall apply:
- Drinking alcohol would be C2H5OH (ethanol), the density of pure alcohol would be 0.9 g/ml
- Assumption: The alcohol is inhaled in the body, i.e. injected with O2.
- Note 1: Calculate first the reaction enthalpy including state of matter
- Note 2: Calculate from ΔHoR and the amount of ethanol per litre of alcohol the amount needed to realize the energy supply
- Note 3: Daily energy consumption is 9200 kJ
- Note 4: for the professionals: the bases of calculation is not fully, in fact, [translation unavailable] the gibbs free energy of reaction at body temperature calculated are. See details of the following exercise thermo-108. But you can see that the result would only change by about 10 percent.
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- Calculs du pH, 4 (SB-62)Un litre de deux solutions A et B avec la même pH de départ de 4.8 est indiqué. A est une solution d'acide salé et B une solution d'acide acide.
Maintenant, on ajoute un litre d'eau à chaque solution.- Calculer le nouveau pH de la solution A et B
- La solution à base de laquelle on a une plus grande modification du pH
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- Forza acida (SB-22)Dato acido acido:CH3COOH{: } H-C<|H><`|H>-C<`||O>-O-H
Quali H's sono acido a il acido acido: coloro che H’s al momento della carbonio o il H al momento della ossigeno?Started by:
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- Calculation of reaction enthalpy 8 (thermo-20)Analogs exercise to thermo-18. It should but only the gibbs free energy of reaction at body temperature for the air conditioning (incineration) of [translation unavailable] calculated are.
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- Calcul titrage 1 (SB-70)Les titrage une acide (20 ml, inconnus concentration) avec une base (0.1 M) nécessaire jusqu'à pour le point d’équivalence 30 ml le base. Comptez les concentrations d'acide à partir de cela.
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- Dichiarazioni, quali corrette (SB-25)Esso è una acque, 1 pioggia soluzione una unprotonico forti acido fabbricato ('HX'). Quale delle seguenti affermazioni è corretta. Scusate la scelta.
- La acido è dissociata da 0.1%
- L'acido è dissociato all'1%
- L'acido è dissociato al 10%
- L'acido è dissociato al 100%
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- Calculation of reaction enthalpy 9 (thermo-21)In a poster roll (diameter 6 cm, length 72 cm) a few drops of pentane (1/4 ml) are given and ignited with the air-pentan mixture.
(a) Calculate the reaction enthalpy.
(b) It has been shown experimentally that too much pentane does not produce a better explosion than the air-pentan mixture Calculate the maximum amount of pentane.
Background: wonderful teaching experience.
Note: unknown size – molecules self to erude.Started by:
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- Calcul QM 1 (qm-01)a) Calculer l'énergie d'un photon de lumière bleue d'une longueur d'onde de 450 nm.
b) Calculer le nombre de particules lumineuses (photons) d'une lampe bleue d'une puissance de 300 watts
Watt est égal au joule/secondeStarted by:
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- Frequenza delle particelle (SB-26)Quali sono le particelle presenti in una soluzione idrica di 1 M H2SO4? In quali quantità relative sono presenti?
pKs(H2SO4) sono -3, pks(HSO–4) = 1.92.
Nota: Es si trovano solo le seguenti particelle: H2O, H3O+, HSO–4, SO2-4, (quasi) nessun H2SO4, poiché l'acido solforico è un acido forteStarted by:
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- Reaction enthalpy, mass difference 1 (thermo-30)In the blast gas reaction 2H + O → H2O, the reaction energy of 571.6 kJ/mol is released.
Note: [J] = [kg m2 s-2]Started by:
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- Calcul QM 2 (qm-02)Les électrons du vent solaire atteignent la Terre après une durée de vol d'environ 30 heures, où ils peuvent produire des aurores boréales.
Les éclairages verts générés par l'excitation de l'oxygène à λ = 557.7 nm sont prédominants dans les lampes polaires.
Il faut calculer si l'énergie cinétique des électrons du vent solaire est suffisante pour stimuler l'oxygène.
Remarque: la distance entre la Terre et le Soleil est de 150 millions de km
Remarque 2: m(e–)= 9.1·10-31 kgStarted by:
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- Interpretazione pKs, 1 (SB-27)Si deve indicare l'acido solforico. il pKs(H2SO4) è -3, pks(HSO–4) = 1.92.
Quale delle seguenti affermazioni è corretta? E quasi più importante: perché?- 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)
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- Reaction enthalpy Mass difference 2 (thermo-31)On [translation unavailable] (https://de.wikipedia.org/w/index.php?title=%C3%84quivalenzvonMasseundEnergie&oldid=245822194) is of which: applications for the mass conversion according to Einstein, that at the same time, heating to 50 degree centigrade of a [translation unavailable] water (adopted: mass 1 tons of) itself a increase in mass to 2.3 micrograms to be given.
Confirm this value with an invoice.Started by:
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- Calcul QM 3 (qm-03)Les estimations (calculs) à partir de, combien de photons une lampe à incandescence avec
100 W par une seconde dans le à la vue région du lumière émis.
Remarque 1: Supposons, pour simplifier, que λ = 500 nmStarted by:
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- Produzione di una determinata concentrazione (SB-40)Descrivi la produzione di queste soluzioni in laboratorio: la soluzione è sempre acqua distillata.
Accettazione le sostanze siano come sali presente e dissolvere si completamente in acqua.- a) 1 litri una 0.3 M NaOH-soluzione
- b) 1 litri una 0.4 M HCL-soluzione
- c) 25 ml di soluzione di 0.5 M di NaOH
- d) 30 ml 0.6 M H2SO4
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- Gibbs free energy of reaction, water [translation unavailable] Hydrogen peroxide (thermo-40)Es reacting H2(g) with O2(g). Assessments with help the free reaction enthalpy whether or not well, you know, H2O(g) or H2O2(g) of which: is. The reaction temperature is 800 Kelvin.
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- Dessinez différents groupes fonctionnels (oc-01)Dessinez à chaque exercice chacun de trois différents molécules, qui ne sont constitués que d'un squelette hydrocarboné (squelette d’alcane, c'est-à-dire liaisons simples) ainsi que du groupe fonctionnel suivant:
- Alcanes non branchées
- Alcanes branchées
- Alcanes cycliques
- Alcées
- Alkyne
- L'alcool
- aldéhyde
- cétone
- Amine
- acide carboxylique
- acide aminé
- ester
- amide
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- Definizione pH (SB-41)Qual è il pH?
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- Calculate seed temperature (thermo-41)Calculate the boiling point of the water .
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- Des dessins différents groupes fonctionnels, 2 (oc-02)Pour chaque exercice dessinez la formule structurelle. En plus du groupe fonctionnel, aucun autre groupe fonctionnel ne peut être dessiné; seuls les liens carbone-hydrogène sont autorisés.
- Un alcool linéaire non branché contenant exactement 3 atomes de carbone
- Un alcool branché avec exactement 5 atomes de carbone
- Toutes les alcools (branchées, non branchées) contenant exactement 4 atomes de carbone
- Trois différentes éther avec chaque au total 4 les atomes de carbone
- Le cétone le moins possible (aussi peu de C que possible)
- L'aldéhyde le moins possible (aussi peu de C que possible)
- Le plus petit possible (le moins de C possible) Amine
- Le plus petit ester possible (aussi peu C que possible)
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- Dessin des groupes fonctionnels différents, 3 (oc-03)Chaque fois, dessinez n'importe quelle molécule qui contient simultanément tous les groupes fonctionnels, la structure de base étant toujours une base d'algues
- Alcène, l'alcool et acide carboxylique
- Alcène, aldéhyde, cétone ainsi que amide
- Acide carboxylique, l'alcool et cétone
- Amin, amide, acide carboxylique et Aldeyhd
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- Il logaritmo (SB-42)Perché si usa il logaritmo?
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- Temperature dependence of equilibrium constant K (thermo-42)Calculate ΔGoR for the combustion reaction temperatures of 298K, 900K and 2300K from N2 to NO. The given temperatures of 900 K correspond to the temperature in an exhaust or in the combustion chamber (2300K),
Note: Calculate the reaction for 1 particle (respectively 1 mol) NO.Started by:
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- Definition of acid (SB-01)What is an acid or a base?
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