A piece of Al(s) at 98°C is placed in 70g of H2O at 26°C. The Aluminum cools and after equilibrium the final temperature of Aluminum and water is 38°C. What was the mass of the aluminum? Cs(H2O)= 4.184J/(g°C) Cs(Al)= 0.90J/(g.°C)

A piece of Al(s) at 98°C is placed in 70g of H2O at 26°C. The Aluminum cools and after equilibrium the final temperature of Aluminum and water is 38°C. What was the mass of the aluminum? Cs(H2O)= 4.184J/(g°C) Cs(Al)= 0.90J/(g.°C)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Calculate the total enthalpy change (kJ) for the following reaction: H2O (s, 90 g, –10. oC) --> H2O (l, 90 g, 10. oC) (The specific heat capacity of ice is 2.0 J/g×oC; the specific heat capacity of liquid water is 4.2 J/g×oC; the enthalpy of fusion for H2O at 0 oC is 6.0 kJ/mol.)
options for all: 1. Reactants. 2. Products.
Below is the reaction for the evaporation of water. H2O (I) H20 (g) AH°f H2O (1)= -285.8 kJ/mole ; H20 (g) = -241.8 kJ/mole %3D Calculate AH*rxn for this process. This is the reaction that happens when your sweat evaporates. Why is this a cooling mechanism for your body? How much sweat would have to evaporate from a person's skin to cool the body by 0.5 C? Assume this person has a body mass of 150 Ibs and the specific heat capacity of the body is 4.0 J/g °C. 1 Ib. = 0.454 kg
Calculate DU for the following reaction at 1 atm and 25oC: CH4 (g) + I2 (g) ----> CH3I (g) + HI (g) DH = +23.0 KJ/mol
Consider the reaction: N2(g) + 3 F2(g) –→ 2 NF3(g) AH° = -249 kJ and AS° = -278 J/K at 25°C Calculate AG° and state whether the reaction is reactant favored or product favored at standard conditions. AG° = -166 kJ; the equilibrium composition should favor reactants. AG° = -332 kJ; the equilibrium composition should favor reactants. AG° = -332 kJ; the equilibrium composition should favor products. None of these AG° = -166 kJ; the equilibrium composition should favor products.
For a particular reaction at 150.5 °C, AG = –392.62 kJ/mol, and AS = 692.39 J/(mol · K). Calculate AG for this reaction at -0.4 °C. - AG = kJ/mol 80 DII DD F3 F4 F7 FB F9 F10 F12 %24 & 4. 5 6 7 8 Y U H K F.
Determine the AU°reaction at 25 °C (in kJ/mol) for the following reaction: 2NO(g) + O2(g)→2NO2(g) Appendix D kJ/mol
For the reaction: N2(g) + 2O2(g) + 16.2 kJ ➜ 2 NO2(g) As nitrogen reacts with oxygen, the heat of formation for nitrogen dioxide is: Select one: a.–8.1 kJ/mole b.+8.1 kJ/mole c.–6.2 kJ/mole d.+16.2 kJ/mole
A piece of Al(s) at 91°C is placed in 69g of H2O at 27°C. The Aluminum cools and after equilibrium the final temperature of Aluminum and water is 35°C. What was the mass of the aluminum?Cs(H2O)= 4.184J/(g°C)Cs(Al)= 0.90J/(g.°C) g
Calculate the standard enthalpy change for the reaction at 25 ∘C. Standard enthalpy of formation values can be found in this list of thermodynamic properties. 2CH3OH(g)+3O2(g)----> 2CO2(g)+4H2O(g)
A 104.4mL sample of 1.8M HNO3 is mixed with a 800mL sample of 0.5M KOH in a Styrofoam cup. If both solutions were initially at 25.2\deg C, and the enthalpy of the neutralization reaction is -57 kJ/mole of H2O formed, what is the final temperature of the mixture? The volume of the formed H2O can be ignored. Assume that the solution has a density of 1.00 g/mL and a specific heat of 4.184 J/g\deg C, and that the Styrofoam cup has an insignificant heat capacity. (enter one decimal place)