% H = 100 - ( 52.14 + 34.73 )=13.13 %
<span>assume 100 g of this compound </span>
<span>mass H = 13.13 g </span>
<span>moles H = 13.13 g / 1.008 g/mol=13 </span>
<span>mass C = 52.14 g </span>
<span>moles C = 52.14 g/ / 12.011 g/mol=4 </span>
<span>mass O = 34.73 g </span>
<span>moles O = 34.73 g/ 15.999 g/mol=2 </span>
<span>the empirical formula is C4H13O2</span>
<u>Answer :</u>
Part 13:
The balanced chemical reaction will be:

Part 14:
The balanced chemical reaction will be:

Part 15:
The balanced chemical reaction will be:

<u>Explanation :</u>
Balanced chemical reaction : It is defined as the reaction in which an individual element of an atom present on reactant side must be equal to product side.
Part 13:
The balanced chemical reaction will be:

Part 14:
The balanced chemical reaction will be:

Part 15:
The balanced chemical reaction will be:

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:
Force acting on the body when the body is at rest the net formals is given