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:
Explanation:
A Bronsted-Lowry base is a substance that accepts a proton in the form of a hydrogen (H) atom.
On the other hand;
Bronsted-Lowry acid is the substance that donates the proton.
HF (aq) + SO32- ⇌ F- + HSO3-
In the forward reaction;
Bronsted-Lowry acid : HF
Bronsted-Lowry base: SO32-
In the backward reaction;
Bronsted-Lowry acid : HSO3-
Bronsted-Lowry base: F-
The conjugate base of HF is F-
The conjugate acid of SO32- is HSO3-
Answer:
About 170-180 grams of potassium nitrate are completely dissolved in 100 g.
Explanation:
Hello!
In this case, according to the reported solubility data for potassium nitrate at different temperatures on the attached picture, it is possible to bear out that about 170-180 grams of potassium nitrate are completely dissolved in 100 g; considering that the solubility is the maximum amount of a solute that can be dissolved in a solvent, in this case water.
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Answer:
0.5 moles of LiOH will absorb 5.6 L of 
Explanation:
According to law of conservation of mass, the sum of mass on the reactant side must be equal to the sum of mass on product side.
The balanced chemical equation is:

2 moles of LiOH absorb 1 mole of
i.e. 22.4 Liters at STP
0.5 moles of LiOH absorb
=
0.5 moles of LiOH will absorb 5.6 L of 
The empirical formula is XeO₃.
<u>Explanation:</u>
Assume 100 g of the compound is present. This changes the percents to grams:
Given mass in g:
Xenon = 73.23 g
Oxygen = 26.77 g
We have to convert it to moles.
Xe = 73.23/
131.293 = 0.56 moles
O = 26.77/ 16 = 1.67 moles
Divide by the lowest value, seeking the smallest whole-number ratio:
Xe = 0.56/ 0.56 = 1
O = 1.67/ 0.56 = 2.9 ≈3
So the empirical formula is XeO₃.