Calculate Percent Error For Melting Point Range
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Melting Point Percent Error Formula
Science & Mathematics Social Science Society & Culture Sports Travel Yahoo Products International Argentina Australia percent error calculator Brazil Canada France Germany India Indonesia Italy Malaysia Mexico New Zealand Philippines Quebec Singapore Taiwan Hong Kong Spain Thailand UK & Ireland Vietnam Espanol About About Answers Community Guidelines Leaderboard Knowledge Partners Points & Levels Blog Safety Tips Science & Mathematics Chemistry Next How calculate percent error for chemistry? I have a test tomorrow in chemistry and i really want to do well. This is one of our problems on our review sheet. Someone please help me find percent error? Marissa determined the melting poin of a substance to be 24.5 degrees Celsius. Find the percent error of her her measurement if the actual melting point is... show more I have a test tomorrow in chemistry and i really want to do well. This is one of our problems on our review sheet. Someone please help me find percent error? Marissa determined the melting poin of a substance to be 24.5 degrees Celsius. Find the percent error of her her measurement if the actual melting point is 31.2 degrees celsius. Please help? I'm really confused. Follow 5 answers 5 Report Abuse Are you sure you want to delete this answer? Yes No Sorry, something has gone wrong. Trending Now Katy Perry Piper Perabo Mark Cuban Witney Carson Free Credit Report Rental Cars Antonio Brown Caelus Energy Buffalo Bills Ford F-150 Answers Relevance Rating Newest Oldest Best Answer: Okay this is fine. take 31.2-24.5 (actual minus thought to be count). you get 6.7 now take 6.7 and divide it by 31.2 and you get an answer now multiply by 100 and that is your error percentage. =21.5% Source(s): Jacob V · 7 years ago 1 Thumbs up 1 Thumbs down Comment Add a comment Submit · just now Report Abuse Calculate Percent Error Source(s): https://shrink.im/a784M catherine · 8 hours ago 0 Thumbs up 0 Thumbs down Comment Add a comment Submit · just now Report Abuse % error = (|Your Result - Accepted Value| / Accepted Value) x 100 so the percent error is the absolute value of what you recorded minus what it should have been. then that is di
inclusion (include_path='.:/usr/lib/php:/usr/local/lib/php') in /home/sciencu9/public_html/wp-content/themes/2012kiddo/header.php on line 46 Science Notes and ProjectsLearn about Science - Do Science Menu Skip to contentHomeRecent PostsAbout Science NotesContact Science NotesPeriodic TablesWallpapersInteractive Periodic TableGrow CrystalsPhysics ProblemsMy Amazon StoreShop Calculate Percent Error 3 Replies Percent error, sometimes referred to as percentage error, is an expression of the difference between a measured value and the known or accepted value. It is often used in science to report the difference between experimental values and expected values.The formula for calculating percent error is:Note: https://answers.yahoo.com/question/?qid=20090923180930AA2Ac3Z occasionally, it is useful to know if the error is positive or negative. If you need to know positive or negative error, this is done by dropping the absolute value brackets in the formula. In most cases, absolute error is fine. For example,, in experiments involving yields in chemical reactions, it is unlikely you will obtain more product http://sciencenotes.org/calculate-percent-error/ than theoretically possible.Steps to calculate the percent error:Subtract the accepted value from the experimental value.Take the absolute value of step 1Divide that answer by the accepted value.Multiply that answer by 100 and add the % symbol to express the answer as a percentage.Now let's try an example problem.You are given a cube of pure copper. You measure the sides of the cube to find the volume and weigh it to find its mass. When you calculate the density using your measurements, you get 8.78 grams/cm3. Copper's accepted density is 8.96 g/cm3. What is your percent error?Solution: experimental value = 8.78 g/cm3 accepted value = 8.96 g/cm3Step 1: Subtract the accepted value from the experimental value.8.96 g/cm3 - 8.78 g/cm3 = -0.18 g/cm3Step 2: Take the absolute value of step 1|-0.18 g/cm3| = 0.18 g/cm3Step 3: Divide that answer by the accepted value.Step 4: Multiply that answer by 100 and add the % symbol to express the answer as a
Characterization of Blue Vitriol CHEMISTRY LAB go back The objective of this lab was to observe certain characteristics of solids especially the melting point. We learned that crystalline solids have a melting point while http://etsoft.tripod.com/old/science.html partial crystals have a melting point range. To determine the melting point, first we had to calibrate the thermometer. We measured the temperature of ice water which is supposed to be zero(0) degrees Celsius. Out thermometer read zero degrees which it was supposed to read. Then we measured the temperature of boiling water which is supposed to be 100 degrees Celsius. Our thermometer measured 100 degrees which was correct. percent error We had to make a boiling point correction by using the formula: BP correction(760mm-atm pressure)*.037c/mm. The number we got was +.074 degrees Celsius which means that water would boil at 100.074 degrees Celsius. This was a negligible difference from 100 degrees so we used the normal boiling point of water. Then we graphed the two measured points against the temperatures we were supposed to get and connected the two points. percent error for This graph can be used to correct the melting point of the unknown. The first unknown we received was lactose(number 26). First we set up a hot water bath and waited to see if the substance would melt below 100 degrees and it didn’t. We had to setup an oil bath to get it to melt. We observed the substance to melt at 224 degrees. The actual melting point range was 222 to 228 degrees. We didn’t find the range for three reasons. First of all we didn’t notice that the substance started to melt at 222 because the temperature was rising too fast. If we removed the heat the temperature would stop rising but once we put the burner back under the oil the temperature would shoot up. Secondly when the temperature reached 224 degrees it looked like the substance melted. The third reason was that maybe the temperature of the oil was rising faster than the thermometer can register it. The real temperature could have been a few degrees higher than what the thermometer read. The percent error we made was from .8% from the low number in the range to 2.7% from the high number. The second unknown we received was sodium thiosulfate(numb
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