Question

bstituting 1.1 M HNO, for 1.1 MHCL O he Repot Sheet, com rid-strong hase reactions. &Change the acid and repeat the neutralization reaci 4. The specifie heat of the metal was determined following Part A of the Esperimental Procedure in this experiment. Complete the fotlowing table for Trial 1 (See Report Sheet) for determining the specific heat of the metal. Record the calculated values with the conrect mumber of significant figures Calculation Zone I. Mass of metal () 2. Temperature of metal ("C) 5 Mass of water (e) 6 Temperature of water ("C 7. Maximum temperature of water ("C) Do 20.94 994 Part 2 A. J00.0 22.0 Ur 24.6 Caleulations for Specifie Heat of a Metal Part 4 1. Temperature change of water, AT (C) 2. Heat gained by water () 3 Show calculation 3. Temperature change of metal, AT (C) 4 4. Specific heat of metal (Mg C) Equation 25.5 Show calculation 5. The enthalpy of solution for the dissolving of a KBr sample was determined following Part C of the Experimental Pro- cedure in this experiment. Complete the following table for Trial 1 (See Report Sheet.) for determining the enthalpy of solution of KBr Calculation Zene 1. Mass of salt (g) 2. Moles of salt (mol) 5.00 Part 2 5. Mass of water (g) 6. Initial temperature of water (C) 7. Final temperature of mixture (C) 25.0 25.0 18.1 Part 4 Calculations for Enthalpy (heat) of Solution for the Dissolution of a Salt 1. Temperature change of solution, ("C) 2. Heat change of water (J) Show calculation, Part 5 3. Heat change of salt (J); sp. ht. of KBr 0.439 J/g "C 4. Total enthalpy change (J) Show calculation 5. AH, (Jmol salt) Equation 25.12 Show calculation =OO Calorimetry

Answer 1

A) Calculations are made:

1) Temperature change:

ΔT = 24.6 - 22 = 2.6 ° C

2) Heat gained by water:

q = m * cp * ΔT = 100 * 4.18 * 2.6 = 1086.8 J

3) Metal temperature change:

ΔT = 24.6 - 99.4 = - 74.8 ° C

4) Specific metal heat:

Cs = - q / m * ΔT = - 1086.8 / 20.94 * -74.8 = 0.694 J / g * ° C

B) Calculations are made:

Mol of salt:

n = g / MM = 5/119 = 0.042 mol

1) Temperature change:

ΔT = 18.1 - 25 = - 6.9 ° C

2) Heat absorbed:

q = 25 * 4.18 * - 6.9 = - 721.05 J

3) Salt heat exchange:

q = 5 * 0.439 * - 6.9 = - 15.15 J

4) Total heat change:

q total = - 721.05 - 15.15 = - 736.2 J

5) Enthalpy change:

ΔH = - q / n = 736.2 / 0.042 = 17528.6 J / mol

If you liked the answer, please rate it in a positive way, you would help me a lot, thank you.