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Math Review Packet Problem solving in chemistry is very important, and for this reason It Is essential to have strong skills in basic math. This is a review packet. It will not be turned in or graded. The purpose of this packet is just to give you more practice in problem solving, An answer key will be posted on moodle. Part 1: Order of Operations: If your class were given the following problem fo simolify, would everyone get the same answer? 30-1272+4-3-5 Some possible answers maybe 34, 16, 4, or 31. . We can arrive at a variety of answers because there is confusion as to which operations should be done in what order. Sometimes symbols of inclusion (parentheses or brackets) are used ta group the numbers together in a specific arrangement to help. To avoid confusion and to assure e that we all get the same answer, we use the order of operations shown below: Order of Operations (PEMDAS is a special acronym to help you remember) 1. Simplify expressions inside grouping symbols. (parentheses) 2. Simplify or evaluate any terms raised to powers. (exponents) 3. Do multiplication and division operations in order as you come to them from left to right. (multiply, divide) 4, Finally, do all addition and subtraction operations in order as you come to them from left to-right. (add, subtract) * Notice the words In the parentheses at thé end of the lines make “pemdas”. Example #1. 30-127244-3-55 Example #2. Since there are no‘parentheses or exponents, using (8 +4) = (10-6) +52 “pemdas” we would begin by doing multiplication and division in order left to right. Leave all other We begin with parentheses: 12+ 4 + & operations alone. Let's group what we need to co. Next evaluate the exponent term 12 + 4 + 25 30 - (12 + 2) +.(4° 3) -5 = 30-6+12-5 Now do multiplication and division in order 3 + 25 Next we will do the addition and subtraction in order left to right. The steps are underlined in order. Finally, add and subtract in order 28 30-64 12-5= 24 + 12 -5= 36 - 5=34 1. Solve each problem below: a. 3(9+1) +B b, 4+5 x (6-2) c. 4+ 10x 23-16 lig) Bo 445 %4 5 41410x Blo b8 3p) 4ra0* U4 BO-llo = 30436 6b] 60 Il d, 21+ 102+ 10 @.10+72-2x5 lOOz1O OFH9-AxS Ae =]31| i0+44-10 =|4Q} 
