Stoich chem reactions practice Answer Section
... ____ 21. For the reaction 2Na + 2H2O 2NaOH + H2, how many grams of sodium hydroxide are produced from 3.0 mol of water? a. 40. g c. 120 g b. 80. g d. 240 g ____ 22. For the reaction 2Na + Cl2 2NaCl, how many grams of chlorine gas are required to react completely with 2.00 mol of sodium? a. 35.5 ...
... ____ 21. For the reaction 2Na + 2H2O 2NaOH + H2, how many grams of sodium hydroxide are produced from 3.0 mol of water? a. 40. g c. 120 g b. 80. g d. 240 g ____ 22. For the reaction 2Na + Cl2 2NaCl, how many grams of chlorine gas are required to react completely with 2.00 mol of sodium? a. 35.5 ...
1984 Advanced Placement Exam
... 38. The radioactive decay of 6C to 7N occurs by the process of 33. The pH of 0.1–molar ammonia is approximately (A) beta particle emission (D) electron capture (A) 1 (B) 4 (C) 7 (D) 11 (E) 14 (B) alpha particle emission (E) neutron capture (C) positron emission 34. …CrO2–+ …OH– …CrO42– + …H2O + …e ...
... 38. The radioactive decay of 6C to 7N occurs by the process of 33. The pH of 0.1–molar ammonia is approximately (A) beta particle emission (D) electron capture (A) 1 (B) 4 (C) 7 (D) 11 (E) 14 (B) alpha particle emission (E) neutron capture (C) positron emission 34. …CrO2–+ …OH– …CrO42– + …H2O + …e ...
SOLUBILITY RULES FOR IONIC COMPOUNDS IN WATER
... 2. Ammonia decomposes according to the reaction: 2NH3 (g) ⇆ N2 (g) + 3H2 (g) A 2.00 liter tank is originally charged with 0.500 moles of ammonia, and at equilibrium it is found that the ammonia is 16.5% decomposed. Calculate the numerical value of the Kc for the above reaction. 3. A tank of O2 has a ...
... 2. Ammonia decomposes according to the reaction: 2NH3 (g) ⇆ N2 (g) + 3H2 (g) A 2.00 liter tank is originally charged with 0.500 moles of ammonia, and at equilibrium it is found that the ammonia is 16.5% decomposed. Calculate the numerical value of the Kc for the above reaction. 3. A tank of O2 has a ...
Chapter 19.1 Balancing Redox Equations
... H2O2 (aq) + 3 I- (aq) + 2 H+ (aq) → 2 H2O (l) + I3- (aq) has an observed rate law rate = k [H2O2] [I-] [H+] A mechanism is proposed for the overall reaction. Which elementary reaction below could be the rate-determining step of the proposed mechanism? a) H2O2 (aq) + 3 I- (aq) + 2 H+ (aq) → 2 H2O (l) ...
... H2O2 (aq) + 3 I- (aq) + 2 H+ (aq) → 2 H2O (l) + I3- (aq) has an observed rate law rate = k [H2O2] [I-] [H+] A mechanism is proposed for the overall reaction. Which elementary reaction below could be the rate-determining step of the proposed mechanism? a) H2O2 (aq) + 3 I- (aq) + 2 H+ (aq) → 2 H2O (l) ...
Document
... Instructions for completion of Section 1 are given on Page 02. SECTION 2 — 80 marks Attempt ALL questions Reference may be made to the Chemistry Higher and Advanced Higher Data Booklet. Write your answers clearly in the spaces provided in this booklet. Additional space for answers and rough work is ...
... Instructions for completion of Section 1 are given on Page 02. SECTION 2 — 80 marks Attempt ALL questions Reference may be made to the Chemistry Higher and Advanced Higher Data Booklet. Write your answers clearly in the spaces provided in this booklet. Additional space for answers and rough work is ...
Chemistry Spell check on
... to the following equation. 2H2O2(aq) → 2H2O(ℓ) + O2(g) (a) At room temperature, the reaction is very slow. It can be speeded up by heating the reaction mixture. State why increasing the temperature causes an increase in reaction rate. ...
... to the following equation. 2H2O2(aq) → 2H2O(ℓ) + O2(g) (a) At room temperature, the reaction is very slow. It can be speeded up by heating the reaction mixture. State why increasing the temperature causes an increase in reaction rate. ...
chemistry intermediate may 2010 marking scheme
... containing the substance D. When a few drops of ammonia solution are added to the solution of D a pale blue precipitate E forms; this precipitate dissolves when more aqueous ammonia is added to form a solution containing the deep blue coloured ion F. (c) G is a colourless volatile liquid with empiri ...
... containing the substance D. When a few drops of ammonia solution are added to the solution of D a pale blue precipitate E forms; this precipitate dissolves when more aqueous ammonia is added to form a solution containing the deep blue coloured ion F. (c) G is a colourless volatile liquid with empiri ...
1994 Released Exam
... Directions: Each set of letteredchoicesbelow refers to the numberedquestionsor statementsimmediately following it. Select the one letteredchoice that bestanswerseach questionor bestfits each statementand then fill in the correspondingoval on the answersheet.A choice may be used once, more than once, ...
... Directions: Each set of letteredchoicesbelow refers to the numberedquestionsor statementsimmediately following it. Select the one letteredchoice that bestanswerseach questionor bestfits each statementand then fill in the correspondingoval on the answersheet.A choice may be used once, more than once, ...
ΔG - Lemon Bay High School
... (a) This process is spontaneous. Whenever two objects at different temperatures are brought into contact, heat is transferred from the hotter object to the colder one. (Section 5.1) Thus, heat is transferred from the hot metal to the cooler water. The final temperature, after the metal and water ach ...
... (a) This process is spontaneous. Whenever two objects at different temperatures are brought into contact, heat is transferred from the hotter object to the colder one. (Section 5.1) Thus, heat is transferred from the hot metal to the cooler water. The final temperature, after the metal and water ach ...
Slide 1
... (a) This process is spontaneous. Whenever two objects at different temperatures are brought into contact, heat is transferred from the hotter object to the colder one. (Section 5.1) Thus, heat is transferred from the hot metal to the cooler water. The final temperature, after the metal and water ach ...
... (a) This process is spontaneous. Whenever two objects at different temperatures are brought into contact, heat is transferred from the hotter object to the colder one. (Section 5.1) Thus, heat is transferred from the hot metal to the cooler water. The final temperature, after the metal and water ach ...
Chemistry - Higher tier - Paper 4 - Sample assessment material
... What is the total volume of gas, measured at room temperature and pressure, made in this reaction? Assume that one mole of gas occupies a volume of 24 dm3 at room temperature and ...
... What is the total volume of gas, measured at room temperature and pressure, made in this reaction? Assume that one mole of gas occupies a volume of 24 dm3 at room temperature and ...
Questions for Study
... standard enthalpy of formation (standard heat of formation) ( H fo ) enthalpy change for the formation of one mole of a substance in its standard state from its elements in their reference forms and in their standard states (6.8) fuel* ...
... standard enthalpy of formation (standard heat of formation) ( H fo ) enthalpy change for the formation of one mole of a substance in its standard state from its elements in their reference forms and in their standard states (6.8) fuel* ...
FM 10-67-2 Chapter 7
... Although secondary standards (substances commonly found in laboratories) can function just as well as a primary standard, they are called secondary standards because of certain characteristics that make them less desirable than primary standards. Foremost among these is the fact that substances typi ...
... Although secondary standards (substances commonly found in laboratories) can function just as well as a primary standard, they are called secondary standards because of certain characteristics that make them less desirable than primary standards. Foremost among these is the fact that substances typi ...
File
... Cu(s) + 2 Ag+ Cu2+ + 2 Ag(s) If the equilibrium constant for the reaction above 36. CuO(s) + H2(g) / Cu(s) + H2O(g) _H = –2.0 kilojoules is 3.71015, which of the following correctly When the substances in the equation above are at describes the standard voltage, E_, and the equilibrium at pressur ...
... Cu(s) + 2 Ag+ Cu2+ + 2 Ag(s) If the equilibrium constant for the reaction above 36. CuO(s) + H2(g) / Cu(s) + H2O(g) _H = –2.0 kilojoules is 3.71015, which of the following correctly When the substances in the equation above are at describes the standard voltage, E_, and the equilibrium at pressur ...
chapter10-bur.320702..
... The negative sign in the above equation occurs because we are measuring the value of q for the surroundings, and qsys = - qsur. If we know the energy of combustion for a compound, in units of kJ/g, then we can say q = m Ucom m = mass of compound burned Ucom = energy of combustion (in kJ/g) Note th ...
... The negative sign in the above equation occurs because we are measuring the value of q for the surroundings, and qsys = - qsur. If we know the energy of combustion for a compound, in units of kJ/g, then we can say q = m Ucom m = mass of compound burned Ucom = energy of combustion (in kJ/g) Note th ...
File
... - in phase equilibrium particles in both phases are gaining or losing kinetic energy, such that they are ________ from one phase to another, while an equal number are moving in the reverse direction e.g. H2O(l) → H20(g) 3. Chemical Reaction Equilibrium - Quantitative reactions are those reactions wh ...
... - in phase equilibrium particles in both phases are gaining or losing kinetic energy, such that they are ________ from one phase to another, while an equal number are moving in the reverse direction e.g. H2O(l) → H20(g) 3. Chemical Reaction Equilibrium - Quantitative reactions are those reactions wh ...
Thermochemistry and calorimetry
... cannot be observed directly owing to the large number of other possible reactions between these two elements. However, the standard enthalpy change for such a reaction be found indirectly from other data, as explained in the next section. • The standard enthalpy of formation of gaseous atoms from th ...
... cannot be observed directly owing to the large number of other possible reactions between these two elements. However, the standard enthalpy change for such a reaction be found indirectly from other data, as explained in the next section. • The standard enthalpy of formation of gaseous atoms from th ...
Thail 1 - University of Missouri
... In the open container, there is no change in volume, ΔH = q However if the products were CO2 and liquid water, but carried out at constant pressure, the enthalpy change for this process however would be –(q + 2(40.65) kJ/mol + 2RT ) ...
... In the open container, there is no change in volume, ΔH = q However if the products were CO2 and liquid water, but carried out at constant pressure, the enthalpy change for this process however would be –(q + 2(40.65) kJ/mol + 2RT ) ...
Lab announcements – 2 lab quiz week before spring break
... Most chemical reactions do not go to completion. chemical equilibrium – two opposing reactions occur simultaneously at the same rate ‘equilibrium’ doesn’t necessarily mean ‘equal’ amounts of reactants and products – in fact, it usually doesn’t. Equilibrium constant – measure of this balance aA + Kc ...
... Most chemical reactions do not go to completion. chemical equilibrium – two opposing reactions occur simultaneously at the same rate ‘equilibrium’ doesn’t necessarily mean ‘equal’ amounts of reactants and products – in fact, it usually doesn’t. Equilibrium constant – measure of this balance aA + Kc ...
Chemistry 400
... 9) A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions were initially at 35.00°C and the temperature of the resulting solution was recorded as 37.00°C, determine the ΔH°rxn (in units of kJ/mol NaOH) for the neutralization r ...
... 9) A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions were initially at 35.00°C and the temperature of the resulting solution was recorded as 37.00°C, determine the ΔH°rxn (in units of kJ/mol NaOH) for the neutralization r ...
Page 1 of 9 Chem 103 Practice Problems: Below is a key for both
... What is the equilibrium constant, K , for the reaction involving the dissolution of silver carbonate in the presence of ammonia: Ag2CO3 (s) + 4NH3 (aq) --> 2Ag(NH3)22+ (aq) + CO32-(aq) K = ? Solution: to get the above equation, need to add the equation (1) + 2 times equation (2). That means that K = ...
... What is the equilibrium constant, K , for the reaction involving the dissolution of silver carbonate in the presence of ammonia: Ag2CO3 (s) + 4NH3 (aq) --> 2Ag(NH3)22+ (aq) + CO32-(aq) K = ? Solution: to get the above equation, need to add the equation (1) + 2 times equation (2). That means that K = ...
equilibrium questions - Southington Public Schools
... Calculate the volume of 0.115 M Ba(OH)2(aq) needed to reach the equivalence point when titrated into a 65.0 mL sample of 0.146 M HOBr(aq). Indicate whether the pH at the equivalence point is less than 7, equal to 7, or greater than 7. Explain. ...
... Calculate the volume of 0.115 M Ba(OH)2(aq) needed to reach the equivalence point when titrated into a 65.0 mL sample of 0.146 M HOBr(aq). Indicate whether the pH at the equivalence point is less than 7, equal to 7, or greater than 7. Explain. ...
Thermometric titration
A thermometric titration is one of a number of instrumental titration techniques where endpoints can be located accurately and precisely without a subjective interpretation on the part of the analyst as to their location. Enthalpy change is arguably the most fundamental and universal property of chemical reactions, so the observation of temperature change is a natural choice in monitoring their progress. It is not a new technique, with possibly the first recognizable thermometric titration method reported early in the 20th century (Bell and Cowell, 1913). In spite of its attractive features, and in spite of the considerable research that has been conducted in the field and a large body of applications that have been developed; it has been until now an under-utilized technique in the critical area of industrial process and quality control. Automated potentiometric titration systems have pre-dominated in this area since the 1970s. With the advent of cheap computers able to handle the powerful thermometric titration software, development has now reached the stage where easy to use automated thermometric titration systems can in many cases offer a superior alternative to potentiometric titrimetry.The applications of thermometric titrimetry discussed on this page are by no means exhaustive. The reader is referred to the bibliography for further reading on the subject.