Percent Error Molar Heat
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Measuring Enthalpy Change Lab Report
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How To Calculate Percent Error In Chemistry
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Enthalpy Change Experiment
Levels Blog Safety Tips Science & Mathematics Chemistry Next Calculate the percentage error in your experimental determination of the molar heat of combustion of magnesium How on earth do I do this. I have been trying for a really long time. Please help. /\H= 1713.3 kJ this is what I got for the molar heat of reaction. So how do enthalpy change of neutralisation experiment I find the % error? Follow 1 answer 1 Report Abuse Are you sure you want to delete this answer? Yes No Sorry, something has gone wrong. Trending Now Answers Best Answer: Percent error=|actual-experimental/actual| x 100% | | is the absolute value sign because it has to be positive. Actual is the accepted/real/theoretical value Experimental is the value you calculated in your experiment. So if the actual was 1800 KJ then you do: | 1800 KJ-1713.3 KJ/1800 KJ | x 100%=4.82% Note*: That is not the actual answer because I don't know the actual value of delta H for that reaction. It's only an example of how you should do it. Source(s): Johnny D · 9 years ago 3 Thumbs up 0 Thumbs down Comment Add a comment Submit · just now Report Abuse Add your answer Calculate the percentage error in your experimental determination of the molar heat of combustion of magnesium How on earth do I do this. I have been trying for a really long time. Please help. /\H= 1713.3 kJ this is what I go
Answers Home All Categories Arts & Humanities Beauty & Style Business & Finance Cars & Transportation Computers & Internet Consumer Electronics Dining Out Education & Reference Entertainment & Music Environment Family & Relationships Food & Drink Games & Recreation Health Home & Garden Local Businesses News & Events Pets Politics & Government Pregnancy & Parenting Science & Mathematics percent error calculator Social Science Society & Culture Sports Travel Yahoo Products International Argentina Australia Brazil Canada France percent error formula Germany India Indonesia Italy Malaysia Mexico New Zealand Philippines Quebec Singapore Taiwan Hong Kong Spain Thailand UK & Ireland Vietnam Espanol About enthalpy change of combustion About Answers Community Guidelines Leaderboard Knowledge Partners Points & Levels Blog Safety Tips Science & Mathematics Chemistry Next Heat of fusion for ice percent error help!? So I got 419 J/C for heat of fusion for https://ca.answers.yahoo.com/question/index?qid=20080329173213AA5WAsm ice. Do I put this over the expected amount to get my percent or under? Idk. Don't ask how I got that. It was an epic fail of a lab. Update: Sorry guys. I noticed I put C instead of gram. Actually, my professor wanted it in J/mole. So it's more like: 7550 J/mole? Update 2: And Yes i calculated that right. i just read it wrong. But yeah. I got 125%. Or should https://answers.yahoo.com/question/index?qid=20111110083025AAqBHZA it be 125%-100%? Update 3: Also, what's the point of factoring in a calorie meter constant to get the average heat of fusion? Update 4: There's a LOT of data. So idk if I can post that right now . The thing is when I i had to get my average of 7550J/mole, I had to factor calorie meter constant of 25.9J/C. I'm wondering if this is because the calorie meter is absorbing some of the energy instead of the ice. Follow 3 answers 3 Report Abuse Are you sure you want to delete this answer? Yes No Sorry, something has gone wrong. Trending Now Shania Twain Bad parenting Gal Gadot Naomie Harris 2016 Crossovers Auto Insurance Quotes Taylor Swift Zahara Pitt Power Rangers Dating Sites Answers Relevance Rating Newest Oldest Best Answer: % error = | measured - accepted| / |accepted| x 100%.. note the absolute value bars!!! % error = | 419 J/g - 334 J/g| / |334 J/g| x 100% = 25%... (2 sig figs) ****** it works out the same right?.. (a - b) / b = (axc - bxc) / (bxc) doesn't it? and to convert from J/g to J/mole.. we just multiply everything by (18g/mole).... that's our "c" :) ******** but if you insist... 419 J/g x (18.02g / mole) = 7
common enthalpy change measurements Enthalpy changes involving solutions The experiments There are a whole range of different enthalpy changes that can be measured by reacting http://www.chemguide.co.uk/physical/energetics/practical.html solutions (or a solution plus a solid) in a simple expanded polystyrene cup. A common example would be the measurement of the enthalpy change of neutralisation of, say, hydrochloric acid and sodium hydroxide solution. The polystyrene cup serves to insulate the reaction mixture, and slows heat losses from the side and bottom. Heat is still lost from the surface of the liquid mixture, percent error of course, and that can be reduced by using a polystyrene lid with a hole for a thermometer. You can allow for heat losses during the reaction by plotting a cooling curve. All of this is described in some detail at the beginning of Chapter 5 of my chemistry calculations book, and I can't repeat it on Chemguide. There are other sources of enthalpy change of error in these experiments - in particular in the accuracy of the thermometer used. Since the temperature rise probably won't be very great, you would need to use the most accurate thermometer possible in order to keep the percentage error low. The calculations The heat evolved or absorbed during a reaction is given by the expression: Heat evolved or absorbed = mass x specific heat x temperature change That can be written in symbols as Heat evolved or absorbed = m s ΔT You will find that the specific heat is sometimes given the symbol "s" and sometimes the symbol "c". The specific heat of a substance is the amount of heat needed to increase the temperature of 1 gram of it by 1 K (which is the same as 1°C). So for water (the value you are most likely to come across), the specific heat is 4.18 J g-1 K-1. That means that it takes 4.18 joules to increase the temperature of 1 gram of water by 1 K (or 1°C). In experiments of this sort, you will quite commonly have measured out a volume of solutio
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