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:
The three laws of Chemical Reaction are .The law of constant proportions. The law of multiple proportions. The law of reciprocal proportions.
A chemical compound is always found to be made up of the same elements combined together in the same fixed proportion by mass.
potassium and chlorine gas ---> chloride.
Hope this helps, have a good day.✌
Answer: For every one mole of Ca used in this reaction, two mols of H20 are used, one mole of Ca(OH)2 is formed, and one mole of H2 is formed.
Explanation: Once the equation is balanced, you can get the ratio from the coefficients. If you are looking at the ratio of Ca to H2O, the ratio is 1:2; Ca to H2 1:1.
Answer:
The answer is a. I learned that they can multiply by using your cells
0.4575 g of saccharine is present in the ten tablets of saccharine dissolved in water.
From the information in the question, the sulfur in saccharin (C7H5NO3S) was completely converted to sulfate ion (SO4^2-). This ion was now reacted with excess barium chloride to form barium sulfate BaSO4. This later reaction occurs as follows;
SO4^2-(aq) + Ba^2+(aq) --------> BaSO4(s)
Number of moles of BaSO4 obtained = 0.5240 g/233 g/mol
= 0.0025 moles
Since the reaction is 1:1, 0.0025 moles of sulfate ions from saccharine reacted.
Molar mass of saccharine = 7(12) + 5(1) + 14 + 3(16) + 32
= 84 + 5 + 14 + 48 + 32 = 183 g/mol
Mass of saccharine in the tablet = 0.0025 moles × 183 g/mol
= 0.4575 g
Learn more: brainly.com/question/1527403
