Compound Î Hrxn (kJ/mol) NH4NO3 + 25.7 KCl + 17.2 NaOH -44.5 KOH -57.6
You are given 6.7 grams of an unknown salt. You dissolve it in 54.1 mL of water. As the salt dissolves, the temperature of the solution changes from an initial temperature of 26.9°C to a final temperature of 53.9°C. Based on the chart of ÎH values and the temperature change, you can determine which salt you have. What is the temperature change? Î T = o C Tries 0/99
This is the ÎT component of the expression: qsurr = C x mass x ÎT where qsurr is the heat lost or gained by the contents of the calorimeter due to the chemical reaction. What is the mass that is changing temperature in this experiment? mass =_______ g
Assume that the heat capacity, C, of the solution is the same as that of water, C = 4.184 J/g oC. Solve for qsurr. qsurr = J Tries 0/99 What is the value of qsys? qsys = ____J
The change in enthalpy, ÎH, of this reaction is equal to qsys, since it was carried out under conditions of constant external pressure. Based on the sign of ÎH (or qsys) the dissolution of the unknown salt is . ________
When 6.7 g of salt are dissolved in 54.1 mL of water, 6880.4 J of heat are given off by the system. How much heat would be given off if only 1.00 g of salt had been dissolved in the water? qsys =_______ J/g
In the table, the ÎH values are given in kJ/mol. Convert your last answer to kJ to give kJ/g. Finally, multiply the kJ/g by the molar masses (g/mol) of possible salts, to get kJ/mol. The reaction of your unknown salt with water was exothermic. So, you will only need to look at salts with negative ÎH values. formula of unknown = ________
Compound Î Hrxn (kJ/mol) NH4NO3 + 25.7 KCl + 17.2 NaOH -44.5 KOH -57.6
You are given 6.7 grams of an unknown salt. You dissolve it in 54.1 mL of water. As the salt dissolves, the temperature of the solution changes from an initial temperature of 26.9°C to a final temperature of 53.9°C. Based on the chart of ÎH values and the temperature change, you can determine which salt you have. What is the temperature change? Î T = o C Tries 0/99
This is the ÎT component of the expression: qsurr = C x mass x ÎT where qsurr is the heat lost or gained by the contents of the calorimeter due to the chemical reaction. What is the mass that is changing temperature in this experiment? mass =_______ g
Assume that the heat capacity, C, of the solution is the same as that of water, C = 4.184 J/g oC. Solve for qsurr. qsurr = J Tries 0/99 What is the value of qsys? qsys = ____J
The change in enthalpy, ÎH, of this reaction is equal to qsys, since it was carried out under conditions of constant external pressure. Based on the sign of ÎH (or qsys) the dissolution of the unknown salt is . ________
When 6.7 g of salt are dissolved in 54.1 mL of water, 6880.4 J of heat are given off by the system. How much heat would be given off if only 1.00 g of salt had been dissolved in the water? qsys =_______ J/g
In the table, the ÎH values are given in kJ/mol. Convert your last answer to kJ to give kJ/g. Finally, multiply the kJ/g by the molar masses (g/mol) of possible salts, to get kJ/mol. The reaction of your unknown salt with water was exothermic. So, you will only need to look at salts with negative ÎH values. formula of unknown = ________
Related textbook solutions
Basic Chemistry
Principles of Chemistry Molecular Approach
Chemistry: Structure and Properties
Principles of Chemistry Molecular Approach
Chemistry: A Molecular Approach
Chemistry: A Molecular Approach
Principles of Chemistry: A Molecular Approach
