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Savatey [412]
3 years ago
9

When the metal was placed in the calorimeter its

Chemistry
2 answers:
liq [111]3 years ago
7 0

Answer:

-79.6C    

-80C  

-81.2C

Explanation:

guajiro [1.7K]3 years ago
5 0

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:

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The density of mercury is 13.6 g/mL What is the mass in kilograms of 5L of mercury?
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What is the molarity of a solution prepared by diluting 250 mL of a 40% H2SO4 solution to 1 Liter? The density of the stock solu
kiruha [24]

Answer:

1.195 M.

Explanation:

  • We can calculate the concentration of the stock solution using the relation:

<em>M = (10Pd)/(molar mass).</em>

Where, M is the molarity of H₂SO₄.

P is the percent of H₂SO₄ (P = 40%).

d is the density of H₂SO₄ (d = 1.17 g/mL).

molar mass of H₂SO₄ = 98 g/mol.

∴ M of stock H₂SO₄ = (10Pd)/(molar mass) = (10)(40%)(1.17 g/mL) / (98 g/mol) = 4.78 M.

  • We have the role that the no. of millimoles of a solution before dilution is equal to the no. of millimoles after dilution.

<em>∴ (MV) before dilution = (MV) after dilution</em>

M before dilution = 4.78 M, V before dilution = 250 mL.

M after dilution = ??? M, V after dilution = 1.0 L = 1000 mL.

∴ M after dilution = (MV) before dilution/(V after dilution) = (4.78 M)(250 mL)/(1000 mL) = 1.195 M.

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3 years ago
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