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Solution Stoichiometry and Chemical Analysis (Section)A

Chemistry: The Central Science | 12th Edition | ISBN: 9780321696724 | Authors: Theodore E. Brown; H. Eugene LeMay; Bruce E. Bursten; Catherine Murphy; Patrick Woodward ISBN: 9780321696724 27

Solution for problem 88E Chapter 4

Chemistry: The Central Science | 12th Edition

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Chemistry: The Central Science | 12th Edition | ISBN: 9780321696724 | Authors: Theodore E. Brown; H. Eugene LeMay; Bruce E. Bursten; Catherine Murphy; Patrick Woodward

Chemistry: The Central Science | 12th Edition

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Problem 88E

Solution Stoichiometry and Chemical Analysis (Section)

A solution is made by mixing 15.0 g of Sr(OH)2 and 55.0 mL of 0.200 M HNO3.

(a) Write a balanced equation for the reaction that occurs between the solutes.

(b) Calculate the concentration of each ion remaining in solution.

(c) Is the resultant solution acidic or basic?

Step-by-Step Solution:
Step 1 of 3

Koshar Amy Brogan March 6 & 8, 2016 Week 13 Cryptography Part 2 Review  Shift Cipher o A code that replaces letters with other letters Code the message “Send Ammo” using the Vignere key “SHOE” (18-7-14-4) S E N D A M M O 18 4 13 3 0 12 12 14 S H O E S H O E 18 7 14 4 18 7 14 4 36 11 27 7 18 19 26 18 10 1 0 K L B H S T A S Now that we have the coded message, try to decode it like as if you received it and needed to read it without knowing what it said. K L B H S T A S 10 11 1 7 18 19 0 18 At this point to work backwards, we need to subtract the code word from what we have, but some of the numbers won’t be big enough sowe have to add 26. K L B H S T A S 10 11 1 7 18 19 0 18 +26 = 36 +26 = 27 +26 = 26 S H O E S H O E -18 -7 -14 -4 -18 -7 -14 -4 18 4 13 3 0 12 12 14 S E N D A M M O Decode this: Key: OUT T I K K U K R Modular Arithmetic  Addition operation o Sum after adding  Multiplication operation o Product after repeated adding  Subtraction operation o Difference after subtracting  Division operation o Quotient after repeated subtracting  Modular Arithmetic o Remainder after dividing With Modular Arithmetic, the answer will be what is left over after dividing the value in question. It doesn’t matter how many timesa number goes into another, only what is left over. With small numbers this is easiest to see using long division Ex 1: 1 = 3 mod2  2 goes into 3 and there is 1 left over (3-2=1) Ex 2: 23 mod5 = 3 20 / 5, 3 left over Ex 3: mod 5 = 0, 1, 2, 3, 4 (all numbers before the value is 5) Ex 4: 27 mod4 = 3 27/4, 3 left over Ex 5: 66 mod11 =0  66 = 11*6 +0 Keep in mind that all of the “=” isn’t reallyan equal sign. It’sreally “≡”, “equivalent”, but I don’t have that symbol in an easy type-able form. With large numbers, it’s easier to put the value in question into a calculator and find out the remainder, but then it will be in decimal form, andwe need it in whole numbers. So: Ex 6: 673 mod 17  673/17 = 39.588  17*39 = 663 673-663 =10 673 mod17 =10 Ex 7: 492 mod 16  492/16 = 30.7516*30 = 480  492-480 = 12  492 mod 16 = 12 Try these: 7 mod 5 = 26 mod 3 = 129 mod4 = Decimation Code In decimation...

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Chapter 4, Problem 88E is Solved
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Textbook: Chemistry: The Central Science
Edition: 12
Author: Theodore E. Brown; H. Eugene LeMay; Bruce E. Bursten; Catherine Murphy; Patrick Woodward
ISBN: 9780321696724

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Solution Stoichiometry and Chemical Analysis (Section)A