The combustion of an organic compound is mostly written as,
CaHbOc + O2 --> CO2 + H2O
where a, b, and c are supposed to be the subscripts of the elements C, H, and O in the compound. Determining the number of moles of C and H in the product which is the same as that in the compound,
(Carbon, C) : (561 mg) x (12/44) = 153 mg x (1 mmole/12 mg) = 12.75
(Hydrogen, H) : (306 mg) x (2/18) = 34 mg x (1 mmole/1 mg) = 34
Calculating for amount of O in the sample,
(oxygen, O) = 255 - 153 mg - 34 mg = 68 mg x (1mmole/16 mg) = 4.25
The empirical formula is therefore,
C(51/4)H34O17/4
C3H8O1
The molar mass of the empirical formula is 60. Therefore, the molecular formula of the compound is,
C9H24O3
Answer:
If there was only 10 hours in a day, including the night time it would be complexly new environment. Say you are up for 7 hours and sleep 3, in those 7 hours you would go to work for probably 4 hours, come home and do stuff for 3 then go to bed and do it all over again.
Newton ,perhaps as many other originations use it as a gravitational law
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:
Answer:
36.09% is the mass percent composition of calcium in calcium chloride
Explanation:
Mass percent is defined as one hundred times the ratio between the mass of a compound (In this case, Calcium), and the total mass of the sample:
<em>Mass Percent = Mass compound / Total mass of the sample × 100</em>
<em />
Computing the values of the problem:
Mass Percent = 0.690g / 1.912g * 100
Mass percent =
<h3>36.09% is the mass percent composition of calcium in calcium chloride</h3>
