April 20, 2024, 11:55:36 AM
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Topic: How to find the average rate based on the average of 2 slopes obtained(Kinetics)  (Read 1975 times)

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Offline Sir Nubs

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Im really stuck on how to find the average rate based on the two slopes i did for my kinetics lab. For our lab we performed 3 experiments and ran 2 trails for each so a total of 6 trials and 6 graphs i have to show. So I converted the trials into seconds and for the first experiment we performed. I graphed the two trials on excel and for tiral 1 of the slope i got y= 0.0501x-1.5321; R2= 0.9977 and for trial two my slope was y= 0.0165x + 0.6452; R2= 0.994. So I know the slopes will be the rate of reaction for each experiment but I know since we want the average of the two slopes in each experiment I need to get 1 answer overall which will equal to having 3 slope values that ill need to plug into to find the determination on m and n. But I dont know how to find the average of the two slopes? I know that I need to do this but i dont know how to find the overall slope. :(

Offline mjc123

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Seriously, you can't find the average of two numbers? Or is there something more complicated that I'm missing here?
Wait a minute. You say "my slope was y= 0.0165x + 0.6452". You are thinking 0.0165x + 0.6542 is the slope? That is the equation of the line. The slope is the multiplier of x, i.e. 0.0165. The other number is the intercept, which you're not interested in for present purposes.
Now, can you work out the average of 0.0165 and 0.0501?

Offline Sir Nubs

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Seriously, you can't find the average of two numbers? Or is there something more complicated that I'm missing here?
Wait a minute. You say "my slope was y= 0.0165x + 0.6452". You are thinking 0.0165x + 0.6542 is the slope? That is the equation of the line. The slope is the multiplier of x, i.e. 0.0165. The other number is the intercept, which you're not interested in for present purposes.
Now, can you work out the average of 0.0165 and 0.0501?
that makes sense. im sorry my lab professor explained it some weird way that I couldnt understand

Offline Babcock_Hall

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Were the conditions of the reaction identical between the two trials, or was something changed?

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