chm151-6solns

chm151-6solns - 6.5 170 kcal 103 cal 4.184 J 1 kJ 3 = 710...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
6.5 170 kcal · 10 3 cal 1 kcal · 4.184 J 1 cal · 1 kJ 10 3 J = 710 kJ The food product with 170 kcal per serving has a greater energy content per serving. 6.7 28.1 J mol K · 1 mol Hg 200.6 g = 0.140 J/g·K 6.16 The final temperature is greater than 22.5 °C, so the 65.1-g sample of water must be warmer than the 108-g sample. q cool water + q warm water = 0 [(108 g)(4.184 J/g·K)(321.1 – 295.7 K)] + [(65.1 g)(4.184 J/g·K)(321.1 K – T i )] = 0 T i = 363 K (90. ºC) 6.19 1.0 × 10 3 mL H 2 O · 1.00 g 1 mL · 333 J/g = 3.3 × 10 5 J 6.26 25.0 mL · 0.80 g 1 mL = 20. g C 6 H 6 q total = energy to cool liquid + energy to change phase from liquid to solid q cool liquid = (20. g)(1.74 J/g·K)(278.7 K – 293.1 K) = –5.0 × 10 2 J q phase change = –(20. g)(127 J/g) = –2500 J q total = –5.0 × 10 2 J + (–2500 J) = –3.0 × 10 3 J (3.0 × 10 3 J released to the surroundings) 6.29 1.00 × 10 3 mL · 0.69 g 1 mL · 1 mol C 8 H 18 114.2 g · 10,922 kJ 2 mol C 8 H 18 = 3.3 × 10 4 kJ heat evolved 6.35 q solution = (155.4 g)(4.2 J/g·K)(289.4 K – 291.8 K) = –1600 J 5.44 g NH 4 NO 3 · 1 mol NH 4 NO 3 80.04 g = 0.0680 mol NH 4 NO 3 q dissolving = – q solution = – Š1600 J 0.0680 mol NH 4 NO 3 · 1 kJ 10 3 J = 23 kJ/mol NH 4 NO 3 6.42 q Pt + q ice = 0 [(9.36 g)( C Pt )(273.2 K – 371.8 K)] + [(0.37 g)(333 J/g)] = 0 C Pt = 0.13 J/g·K
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
6.43
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 5

chm151-6solns - 6.5 170 kcal 103 cal 4.184 J 1 kJ 3 = 710...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online