Ba (OH)2 (s) + 2 NH4Cl (s) → BaCl2 (s) + 2 NH3 (g) + 2 H2O (l)
DHf° (kJ/mol) -626.56 -314.55 -495.73 -45.90 -285.83
Mass of water | 5.678 g |
Ti | 2.33 |
Tf | -21.33 |
a. calculate the heat lost when the mass of water (g) cools from Ti to 0 °C. This is q1. This is the change in part A of the graph.
b. calculate the heat lost when the mass of water changes from liquid to solid (ice). This is q2. This is the change in part B of the graph.
c. calculate the heat lost when the water cools from 0 °C to Tf. This is q3. This is the change in part C of the graph.
d. calculate the total heat lost from the surroundings: q1 + q2 + q3.
e. calculate the heat absorbed by the system.
f. calculate the value of DHrxn° for the reaction in the system.
g. calculate the number of moles of barium hydroxide in a reaction that absorbs the amount of heat you calculated in 5.
h. calculate the number of moles of ammonium chloride required to react with the mass of barium hydroxide calculated in 7.
I. calculate the grams of each reactant required.
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would you please review my answers and help me with the ones
that not answered because im stuck. questions 1-9
the mass of water us 3.549g
the initial temp of water is 12.19 degree C
the final temperature of the water is -14.01 degree C
How much of each reactant do you need to rernove enough heat to freeze the water? That is the question you are going to answer. A The system, surroundings and the flow of beat In...
can you please help me with this. im really confused specially
on question 2
1. Voice the mass of water (g) 3549g the initial temperature of the water (7 the fmal temperature of the ice ( 12.19 C 14.01 C You will then calculate 1. the heat lost when the mass of water (g) cools from Ti to 0°C. This is i. This is the change in part A of the graph. 2. the heat lost when the mass of...
Part A: Calibration of Calorimeter 1. Calculate the heat lost by your hot water (q=mx Cs x (Tf-Ti) m: 50.0 g Tf=24.1 C Ti=30.6 2. Calculate the heat gained by your cold water m=50.0g Tf=24.1 C Ti=19.1 C 3. The difference in the heat lost by the hot water and that gained by the cool water is the heat gained by the calorimeter. Calculate the heat gained by the calorimeter here. 4. Now calculate the heat capacity of your calorimeter....
Test Object Object Ti (°C) Water Ti (°C) Object + Water Tf (°C) Water DT = Tf - Ti (°C) Object DT = Tf - Ti (°C) Object mass (g) Water Mass (g) Steel bolt 81 C 20 C 36 C 16 C 65 C 46 g 150 g The heat lost by the hot bolt is equal to the heat gained by the water in the calorimeter. Use the equations provided in the eScience manual and what you know...
It can be shown that as a mass m with specific heat c changes temperature from Ti to Tf its change in entropy is ΔS=mcln(Tf/Ti) if the temperatures are expressed in kelvin. Suppose you put 78 g of milk at 278 K into an insulated cup containing 290 g of coffee at 355 K, and that each has the specific heat of water. The system comes to an equilibrium temperature of 339 K. Part A What is the entropy change...
Calorimetry Lab
-Finding speficic heat capacity of calorimeter C(cal)
I need this by today pls help!
Exp.5 CHEMICAL EQUILIBRI 2. (2 marks) Table 1: Determining the heat capacity of the calorimeter. Run 1 Run 2 Run 3 Mass of the hot water, m. (g): Initial temperature of the hot water, T. (°C): Mass of the cold water, me (g): Initial temperature of the cold water and calorimeter, T. (°C): Final temperature of the mixture, T. (°C): Heat lost by the...
1) The heat of solution (delta H) For sodium hydroxide is
-44.5 kJ/mol calculate the amount of energy involved when 5.0 g
sodium hydroxide is dissolved in water
2) calculate the change in temperature expected when 5.0 g
sodium hydroxide is dissolved in 50.0 g water using the energy
(Joules) calculated above. (Ccal= 4.5J/g°C, include 4.0g magnetic
stir-bar in the total mass)
Prelab Exercise: The Heat of solution (H) for sodium hydroxide is -44.5 kJ/mol. Calculate the amount of energy...
use the heat equation to calculate the energy in joules and calories for each of the following. A lost when 75.8g of water cools from 86.4°c to 3.1°c. B Lost when 75.8 g of water cools from 86.4°c to 3.1°c
An ice cube of mass 500 g at 0 °C is dropped into an insulated container of 1.0 kg of water that initially is at room temperature (25 °C), and eventually the system reaches equilibrium. The insulator is not perfect, so 20 kJ of heat flows from the room into the water during the process. 3. a. Calculate the entropy increase in the ice that melts into water. b. Calculate the entropy loss of the water that cools down. c....
In the diagram q1 =15.0μC and q2 = - 2.0μC are placed on the two
corners of an equilateral triangle abc of base L = 3.0cm.
a. What is the electric potential energy (in units of
J) of q1 and
q2?
b. What is the electric potential (in units of
106Volts) at c due to
q1 and q2?
c. Now, you bring a particle of mass m=1.5
x10-6kg and excess charge q3 =
5.0 μC from infinity and place it...