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Chapter 4,5,6
... reaction is a redox reaction indicate which species is oxidized, which is reduced, and how many moles of electrons are transfered for each mol of reaction. (a) Cu (s) + 2 Ag(NO3) (aq) → 2 Ag (s) + Cu(NO3)2 (aq) (b) HCl (g) + NH3 (g) → NH4Cl (s) (c) 16 H+ (aq) + 2 Cr2O72- (aq) + C2H6O (l) → 4 Cr3+ (a ...
... reaction is a redox reaction indicate which species is oxidized, which is reduced, and how many moles of electrons are transfered for each mol of reaction. (a) Cu (s) + 2 Ag(NO3) (aq) → 2 Ag (s) + Cu(NO3)2 (aq) (b) HCl (g) + NH3 (g) → NH4Cl (s) (c) 16 H+ (aq) + 2 Cr2O72- (aq) + C2H6O (l) → 4 Cr3+ (a ...
CH2 Student Revision Guides pdf
... The electronegativity index is a measure of how strongly an atom in a compound attracts the pair of electrons in a bond. Pauling gave values for the electronegativity index and some values are shown below. H ...
... The electronegativity index is a measure of how strongly an atom in a compound attracts the pair of electrons in a bond. Pauling gave values for the electronegativity index and some values are shown below. H ...
Experiment 9
... IONIC AND HETEROGENEOUS EQUILIBRIA IN SOLUTIONS Some substances while being dissolved interact with molecules of a solvent. As a result they dissociate and form ions. The process of dissociation can be written as: ...
... IONIC AND HETEROGENEOUS EQUILIBRIA IN SOLUTIONS Some substances while being dissolved interact with molecules of a solvent. As a result they dissociate and form ions. The process of dissociation can be written as: ...
ATOMS, MOLES AND STOICHIOMETRY
... Ion formed/ name Cu+, copper (I), Cu2+, copper (II) Zn2+, Zinc Fe2+, Iron (II), Fe3+, Iron (III) Mn2+, Manganese (II) Cr3+, Chromium (III) Ag+, Silver Learn these!! ...
... Ion formed/ name Cu+, copper (I), Cu2+, copper (II) Zn2+, Zinc Fe2+, Iron (II), Fe3+, Iron (III) Mn2+, Manganese (II) Cr3+, Chromium (III) Ag+, Silver Learn these!! ...
Gas Laws
... 4. A mixture in which the particles are so small that they will not reflect the “light” from a laser are called ________. 5. A solution that contains a large amount of solute per amount of solvent is called a _____________ solution. 6. What is the term that is used to describe a solid being formed w ...
... 4. A mixture in which the particles are so small that they will not reflect the “light” from a laser are called ________. 5. A solution that contains a large amount of solute per amount of solvent is called a _____________ solution. 6. What is the term that is used to describe a solid being formed w ...
File
... The reactivity of alkali metals decreases going down the group. What is the reason for this? The atoms of each element get F larger going down the group. This means that the outer shell gets further away from the nucleus and is shielded by more electron shells. Cl The further the outer shell ...
... The reactivity of alkali metals decreases going down the group. What is the reason for this? The atoms of each element get F larger going down the group. This means that the outer shell gets further away from the nucleus and is shielded by more electron shells. Cl The further the outer shell ...
2013-2014
... Section A consists of multiple-choice question in this question paper, while Section B contains conventional questions printed separately in Question-Answer Book B. Answers to Section A should be marked on the Multiple-choice Answer Sheet while answers to Section B should be written in the spaces pr ...
... Section A consists of multiple-choice question in this question paper, while Section B contains conventional questions printed separately in Question-Answer Book B. Answers to Section A should be marked on the Multiple-choice Answer Sheet while answers to Section B should be written in the spaces pr ...
1 R R 1Ch Ro_ R___ + ____ ____ + _+ S ___y → +
... 12. Potassium chloride + Silver nitrate → 13. Calcium hydroxide + Hydrogen nitrate → 14. Lead II nitrate + Potassium chloride Æ 15. Strontium carbonate + Hydrogen nitrate Æ 16. Gold + Potassium nitrate Æ 17. Zinc + Silver nitrate Æ 18. Aluminum + Copper II sulfate Æ ...
... 12. Potassium chloride + Silver nitrate → 13. Calcium hydroxide + Hydrogen nitrate → 14. Lead II nitrate + Potassium chloride Æ 15. Strontium carbonate + Hydrogen nitrate Æ 16. Gold + Potassium nitrate Æ 17. Zinc + Silver nitrate Æ 18. Aluminum + Copper II sulfate Æ ...
Chapter 8 and 9 homework
... added to aqueous sodium sulfide to produce solid iron(III) sulfide and aqueous sodium sulfate. 15. What mass of Li3PO4 is needed to prepare 500.0 mL of a solution having a lithium ion concentration of 0.175 M? 16. 17.5 mL of a 0.1050 M Na2CO3 solution is added to 46.0 mL of 0.1250 M NaCl. What is th ...
... added to aqueous sodium sulfide to produce solid iron(III) sulfide and aqueous sodium sulfate. 15. What mass of Li3PO4 is needed to prepare 500.0 mL of a solution having a lithium ion concentration of 0.175 M? 16. 17.5 mL of a 0.1050 M Na2CO3 solution is added to 46.0 mL of 0.1250 M NaCl. What is th ...
AP 2005 Chemistry Free-Response Questions
... one end that has been ignited and extinguished, but still contains hot, glowing, partially burned wood). (b) The following three mixtures have been prepared: CaO plus water, SiO2 plus water, and CO2 plus water. For each mixture, predict whether the pH is less than 7, equal to 7, or greater than 7. J ...
... one end that has been ignited and extinguished, but still contains hot, glowing, partially burned wood). (b) The following three mixtures have been prepared: CaO plus water, SiO2 plus water, and CO2 plus water. For each mixture, predict whether the pH is less than 7, equal to 7, or greater than 7. J ...
Chemistry 30 Review of Basic Chemistry 20
... named, use brackets to keep that complex ion as a group. ...
... named, use brackets to keep that complex ion as a group. ...
Time
... The Ksp for silver iodide, AgI is known to be 1.5 x 10-16. Determine the [Ag+] in a saturated solution of silver iodide. ...
... The Ksp for silver iodide, AgI is known to be 1.5 x 10-16. Determine the [Ag+] in a saturated solution of silver iodide. ...
General Chemistry Questions
... 7. Consider the following gas-phase equilibrium: H2(g) + I2(g) ↔ 2HI(g) At a certain temperature, the equilibrium constant Kc is 4.0. Starting with equimolar quantities of H2 and I2 and no HI, when equilibrium was established, 0.20 moles of HI was present. How much H2 was used to start the reaction ...
... 7. Consider the following gas-phase equilibrium: H2(g) + I2(g) ↔ 2HI(g) At a certain temperature, the equilibrium constant Kc is 4.0. Starting with equimolar quantities of H2 and I2 and no HI, when equilibrium was established, 0.20 moles of HI was present. How much H2 was used to start the reaction ...
Ch 8 Lecture Notes
... Sample Exercise 8.7 – Prediction the Mass of a Precipitate Barium sulfate is used to enhance Xray imaging of the upper and lower GI tract. In upper GI imaging, patients drink a suspension of solid barium sulfate in H 2O. The compound is not toxic because of its limited solubility. To make pure bari ...
... Sample Exercise 8.7 – Prediction the Mass of a Precipitate Barium sulfate is used to enhance Xray imaging of the upper and lower GI tract. In upper GI imaging, patients drink a suspension of solid barium sulfate in H 2O. The compound is not toxic because of its limited solubility. To make pure bari ...
I- Introduction
... The Scope of Analytical Chemistry: Analytical chemistry has bounds which are amongst the widest of any technological discipline. An analyst must be able to design, carry out, and interpret his measurements within the context of the fundamental technological problem with which he is presented. The se ...
... The Scope of Analytical Chemistry: Analytical chemistry has bounds which are amongst the widest of any technological discipline. An analyst must be able to design, carry out, and interpret his measurements within the context of the fundamental technological problem with which he is presented. The se ...
Lecture 4
... Reactions that result in the formation of an insoluble product are known as precipitation reactions. (See figure I.1) A precipitate is an insoluble solid formed by a reaction in solution. ...
... Reactions that result in the formation of an insoluble product are known as precipitation reactions. (See figure I.1) A precipitate is an insoluble solid formed by a reaction in solution. ...
Pre-AP Chemistry - Simple Rules for Electron Exchange Simple
... Simple rules for assigning oxidation numbers Tracking electron gain and loss for simple reactions like metals becoming ionized is easy. But in most chemical reactions, it is impossible to simply look at the reactants and products and track the electron exchange. In those cases, we must do some elect ...
... Simple rules for assigning oxidation numbers Tracking electron gain and loss for simple reactions like metals becoming ionized is easy. But in most chemical reactions, it is impossible to simply look at the reactants and products and track the electron exchange. In those cases, we must do some elect ...
acid
... “insoluble” compounds will dissolve to a slight extent. For “soluble” compounds there will be a limit as to the amount of compound that will dissolve in a given amount of water. We can also discuss solubility in other solvents. ...
... “insoluble” compounds will dissolve to a slight extent. For “soluble” compounds there will be a limit as to the amount of compound that will dissolve in a given amount of water. We can also discuss solubility in other solvents. ...
Chemistry Lesson Plans #07 - Chemical Reactions
... usually true • One product is only slightly soluble and precipitates from solution. For example sodium sulfide reacts with cadmium nitrate to produce a yellow precipitate of cadmium sulfide in a solution of sodium nitrate: o Na2 S (aq ) + Cd ( NO3 ) 2 (aq ) → CdS ( s ) + 2 NaNO2 (aq) ...
... usually true • One product is only slightly soluble and precipitates from solution. For example sodium sulfide reacts with cadmium nitrate to produce a yellow precipitate of cadmium sulfide in a solution of sodium nitrate: o Na2 S (aq ) + Cd ( NO3 ) 2 (aq ) → CdS ( s ) + 2 NaNO2 (aq) ...
Question paper - Unit A173/02 - Module C7 - Higher tier
... Use black ink. HB pencil may be used for graphs and diagrams only. Answer all the questions. Read each question carefully. Make sure you know what you have to do before starting your answer. Write your answer to each question in the space provided. Additional paper may be used if necessary but you m ...
... Use black ink. HB pencil may be used for graphs and diagrams only. Answer all the questions. Read each question carefully. Make sure you know what you have to do before starting your answer. Write your answer to each question in the space provided. Additional paper may be used if necessary but you m ...
Day 13 Main Group Pt 1
... A. Introduction. The alkali metals and alkaline earth metals both provided us with an impression of the general similarities that exist within each group. Differences were subtle. Starting with Group 3, the differences within each group are more striking than the similarities. For example in Group I ...
... A. Introduction. The alkali metals and alkaline earth metals both provided us with an impression of the general similarities that exist within each group. Differences were subtle. Starting with Group 3, the differences within each group are more striking than the similarities. For example in Group I ...
CHM 1033 Chemistry for Health Sciences
... 22. How much heat is required to warm 2.00 liters of water from 25.0 °C to 95.0 °C? (The density of water is 1.00 g/mL) See data on Reference notes (*) at the end of this section. (a) 5.86 x 105 kJ (b) 1.40 x 102 kJ (c) 3.35 x 101 kJ (d) 5.86 x 102 kJ (e) -1.40 x 102 kJ 23. Suppose that 5.00 g of si ...
... 22. How much heat is required to warm 2.00 liters of water from 25.0 °C to 95.0 °C? (The density of water is 1.00 g/mL) See data on Reference notes (*) at the end of this section. (a) 5.86 x 105 kJ (b) 1.40 x 102 kJ (c) 3.35 x 101 kJ (d) 5.86 x 102 kJ (e) -1.40 x 102 kJ 23. Suppose that 5.00 g of si ...
Experimental skills and abilities
... 1 The evaporation process should be done very slowly. This is because sugar can easily char as it solidifies around the sides of the evaporating basin during the evaporating process. Also the crystallisation will require a lot longer for crystals to form from the concentrated solution and may need ...
... 1 The evaporation process should be done very slowly. This is because sugar can easily char as it solidifies around the sides of the evaporating basin during the evaporating process. Also the crystallisation will require a lot longer for crystals to form from the concentrated solution and may need ...
Hydroxide
Hydroxide is a diatomic anion with chemical formula OH−. It consists of an oxygen and hydrogen atom held together by a covalent bond, and carries a negative electric charge. It is an important but usually minor constituent of water. It functions as a base, a ligand, a nucleophile and a catalyst. The hydroxide ion forms salts, some of which dissociate in aqueous solution, liberating solvated hydroxide ions. Sodium hydroxide is a multi-million-ton per annum commodity chemical. A hydroxide attached to a strongly electropositive center may itself ionize, liberating a hydrogen cation (H+), making the parent compound an acid.The corresponding electrically neutral compound •HO is the hydroxyl radical. The corresponding covalently-bound group -OH of atoms is the hydroxyl group.Hydroxide ion and hydroxyl group are nucleophiles and can act as a catalyst in organic chemistry.Many inorganic substances which bear the word ""hydroxide"" in their names are not ionic compounds of the hydroxide ion, but covalent compounds which contain hydroxyl groups.