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Bumek [7]
3 years ago
15

Calculate the heat change involved when 2.00 L of water is heated from 20.0/C to 99.7/C in

Chemistry
1 answer:
IgorC [24]3 years ago
7 0

Answer:

669.48 kJ

Explanation:

According to the question, we are required to determine the heat change involved.

We know that, heat change is given by the formula;

Heat change = Mass × change in temperature × Specific heat

In this case;

Change in temperature = Final temp - initial temp

                                       = 99.7°C - 20°C

                                       = 79.7° C

Mass of water is 2000 g ( 2000 mL × 1 g/mL)

Specific heat of water is 4.2 J/g°C

Therefore;

Heat change = 2000 g × 79.7 °C × 4.2 J/g°C

                      = 669,480 joules

But, 1 kJ = 1000 J

Therefore, heat change is 669.48 kJ

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Answer:

Explanation:

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3 years ago
When the metal was placed in the calorimeter its
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Answer:

Exam 3 Material

Homework Page Without Visible Answers

This page has all of the required homework for the material covered in the third exam of the first semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.

Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.

These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.

Explanation:

5 0
2 years ago
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3 years ago
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A silver cube with an edge length of 2.42 cm and a gold cube with an edge length of 2.75 cm are both heated to 85.4 ∘C and place
kakasveta [241]

Answer:

Explanation:

Volume of silver cube = 2.42³ = 14.17 cm³

mass of silver cube = volume x density

= 14.17 x 10.49 = 148.64 gm

Volume of gold cube = 2.75³ = 20.8  cm³

mass of gold cube =  20.8 x 19.3 = 401.44 gm

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Heat absorbed = heat lost = mass x specific heat x temperature fall or rise

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= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )

= (35.67 + 51.78 ) x ( 85.4 - T )

87.45 x ( 85.4 - T )

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7 0
3 years ago
Explain how this lab simulates the various isotopes of an element
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A trustable source was above to say that the first experiments will decide how many isotopes of this element exist. Isotopes are atoms of an element which bear chemically the identical but have dissimilar physical properties. One adaptable property among isotopes is their atomic mass.
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