i Chrome File Edit View History Bookmarks People Tab Window HelpChem101O General Chemistry I (LAB SC x +* 8%D Sat 1:10 PM Q OEi app. 101edu.co10 Less Known Sleep Diswww.youtube.comRecommended Psych2GoCloseE Apps MM Gmail D YouTube Maps Translate 3 https://www.carth.SettingsO Google Chrome isn't your default browserSet as defaultQuestion 8 of 8SubmitA 130.3 g piece of copper (specific heat 0.38 Jİ •°C) is heated andthen placed into 400.0 g of water initially at 20.7°C. The waterincreases in temperature to 22.2°C. What is the initial temperature of°Cthe copper? (The specific heat of water is 4.18 J/g.°C).1346C89+/-х 10010,058dtv24MacBook Air80esc$&%23delete46789.13PQWERтYFGJKreturADs lock>VNMz'HcommandoptionoptioncommandcontrolR 3322.

We know that, according to the law of conservation of heat energy, energy gained or loss by the…

A 130.3 g piece of copper (specific heat 0.38 Jig. C) is heated and then placed into 400.0 g of water initially at 20.7°C. The water increases in temperature to 22.2°C. What is the initial temperature of the copper? (The specific heat of water is 4.18 J/g C).

A 130.3 g piece of copper (specific heat 0.38 Jig. C) is heated and then placed into 400.0 g of water initially at 20.7°C. The water increases in temperature to 22.2°C. What is the initial temperature of the copper? (The specific heat of water is 4.18 J/g C).

Appl Of Ms Excel In Analytical Chemistry

2nd Edition

ISBN:9781285686691

Author:Crouch

Publisher:Crouch

Chapter2: Basic Statistical Analysis With Excel

Section: Chapter Questions

Problem 12P

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Question 8 of 8
Submit
A 130.3 g piece of copper (specific heat 0.38 Jİ •°C) is heated and
then placed into 400.0 g of water initially at 20.7°C. The water
increases in temperature to 22.2°C. What is the initial temperature of
°C
the copper? (The specific heat of water is 4.18 J/g.°C).
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MacBook Air
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delete
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8
9.
1
3
P
Q
W
E
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K
retur
A
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Chemistry (i) For T = 298K ΔG° = -8.314 x 298 x ln (2.83x10-3) = 14.524 KJ/mol (ii) For T = 308.15K ΔG° = -8.314 x 308.15 x ln (9.619x10-3) = 11.891 KJ/mol (iii) For T = 318.15K ΔG° = -8.314 x 318.15 x ln (3.14x10-2) = 9.147 KJ/mol (iv) For T = 328.15K ΔG° = -8.314 x 328.15 x ln (8.82x10-2) = 6.621 KJ/mol (v) For T = 338.15K ΔG° = -8.314 x 338.15 x ln (2.42x10-1) = 3.990 KJ/mol Based on the measured ΔG° values, is this equilibrium spontaneous at room temperature? Which factor, entropy, or enthalpy, has the greater impact on spontaneity in this case? Explain your answers.
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