Foundations of College Chemistry 15e Binder Ready Version + WileyPLUS Registration Card
Foundations of College Chemistry 15e Binder Ready Version + WileyPLUS Registration Card
15th Edition
ISBN: 9781119231318
Author: Morris Hein
Publisher: Wiley (WileyPLUS Products)
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Chapter 5, Problem 5PE
Interpretation Introduction

Interpretation:

The general arrangement of subatomic particles in the atom has to be described.

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5 What would the complete ionic reaction be if aqueous solutions of potassium sulfate and barium acetate were mixed? ed of Select one: O a 2 K SO4 + Ba2 +2 C₂H3O21 K+SO4 + Ba2+ + 2 C2H3O21 K+SO42 + Ba2 +2 C2H3O2 BaSO4 +2 K+ + 2 C2H3O estion Ob. O c. Od. 2 K SO4 +Ba2 +2 C₂H₂O₂ BaSO4 + K+ + 2 C2H3O BaSO4 + K + 2 C2H301 →Ba² +SO42 +2 KC2H3O s page
(28 pts.) 7. Propose a synthesis for each of the following transformations. You must include the reagents and product(s) for each step to receive full credit. The number of steps is provided. (OC 4) 4 steps 4 steps OH b.
LTS Solid: AT=Te-Ti Trial 1 Trial 2 Trial 3 Average ΔΗ Mass water, g 24.096 23.976 23.975 Moles of solid, mol 0.01763 001767 0101781 Temp. change, °C 2.9°C 11700 2.0°C Heat of reaction, J -292.37J -170.473 -193.26J AH, kJ/mole 16.58K 9.647 kJ 10.85 kr 16.58K59.64701 KJ mol 12.35k Minimum AS, J/mol K 41.582 mol-k Remember: q = mCsAT (m = mass of water, Cs=4.184J/g°C) & qsin =-qrxn & Show your calculations for: AH in J and then in kJ/mole for Trial 1: qa (24.0969)(4.1845/g) (-2.9°C)=-292.37J qsin = qrxn = 292.35 292.37J AH in J = 292.375 0.2923kJ 0.01763m01 =1.65×107 AH in kJ/mol = = 16.58K 0.01763mol mol qrx Minimum AS in J/mol K (Hint: use the average initial temperature of the three trials, con Kelvin.) AS=AHIT (1.65×10(9.64×103) + (1.0 Jimai
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