Any hard facts on breathing e-liquid causing nic levels to drop?

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5bucks

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I'm done ordering juice with nicotine as I am over my addiction. I have a lot of 6 mg and 11 mg juice that I don't want to waste. I've seen Rip Trippers on YouTube say that leaving the caps off the bottles diminishes the NIC content. Has anyone done any hard science to confirm this? Maybe someone with a NIC testing kit? I'll leave the caps off my juice if it is in fact the case.

Thanks!
 

Cullin Kin

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When he says practical he means that nicotine does not decay quickly, if at all. Nicotine tends to oxidize and gain a funk in e juice, but you will be waiting a very very long time. By the time even your 6 mg juice decays to zero (which could be never), your PG and VG will have gone so bad you will want nothing to do with the juice. If you choose to vape nasty moldy bacteria ridden juice, that's up to you.

Edit: I just asked my chemical engineering professor and she confirmed.

People have left nicotine juice for DIY out for 5 years and it has maybe lost 1 mg of it's overall nicotine content. Then again, that may just be whatever device they used to measure it.

Consider this, three year old cigarettes are stale, beyond stale, but the nicotine is just as active as it was when they were first made.

Best options: Sell it or give it away.
 
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Cullin Kin

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OK. So no hard science then.

Thanks for the unsolicited opinions on what I should do with my juice. :/

What would you like? A decay rate algorithm? And exponential decay model? Because there isn't.

Here, let's try. Our base population model: N(t) = Ne^(kt) Here we are looking for a decay model so k will be negative making our model N(t) = Ne^(-kt). We know someone may have lost 1 mg of 100 mg nicotine juice in 5 years. Let's be generous and say it was 1mg/mL in lost potency which it was not, I just want to show you how long this would take even with that much lost. That means after 5 years we still have 99% of our potency left. No matter the starting amount we can set up our model like this: 0.99N = Ne^(k(5)) The N's can cancel and we are left with 0.99 = e^(k(5)) Taking the natural log of each side we get: ln(0.99) = 5k and finally our decay rate: k = ln(0.99)/5 ~ -0.00201007 We now have out complete model with a decay rate: N(t) = Ne^(-0.00201007t) with t in years. Let's figure out the half-life if nicotine with this model.

0.5N = Ne^(-0.00201007t) canceling the N's and taking the natural log of each side we get: ln(0.5) = -0.00201007t therefor the time for even half of any amount of nicotine to remain would be: t = ln(0.5)/(-0.00201007) ~ 344.84 Years.

This is assuming nicotine decays at an exponential rate. It could be linear, logistic, who knows.

This also assumes that whatever device this person used to measure nicotine was any sort of accurate and precise.

The point being, there is no hard science of the decay rate of nicotine in e-juice. Just try a google search, nothing but pure speculation.

I wish you good luck.
 
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Cullin Kin

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Welcome to ECF! Might sound a bit picky but really your header is kind of off. You don't "breath" nic... the word is "inhale" and, truly, you absorb it through skin and membranes making it unnecessary to inhale it at all... the mouth absorption is just fine as any pipe, cigar smoker or dipper knows.

I think he means breathing like taking the top off a cap of juice and exposing it to air for a period of time.
 

5bucks

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What would you like? A decay rate algorithm? And exponential decay model? Because there isn't.

Here, let's try. Our base population model: N(t) = Ne^(kt) Here we are looking for a decay model so k will be negative making our model N(t) = Ne^(-kt). We know someone may have lost 1 mg of 100 mg nicotine juice in 5 years. Let's be generous and say it was 1mg/mL in lost potency which it was not, I just want to show you how long this would take even with that much lost. That means after 5 years we still have 99% of our potency left. No matter the starting amount we can set up our model like this: 0.99N = Ne^(k(5)) The N's can cancel and we are left with 0.99 = e^(k(5)) Taking the natural log of each side we get: ln(0.99) = 5k and finally our decay rate: k = ln(0.99)/5 ~ -0.00201007 We now have out complete model with a decay rate: N(t) = Ne^(-0.00201007t) with t in years. Let's figure out the half-life if nicotine with this model.

0.5N = Ne^(-0.00201007t) canceling the N's and taking the natural log of each side we get: ln(0.5) = -0.00201007t therefor the time for even half of any amount of nicotine to remain would be: t = ln(0.5)/(-0.00201007) ~ 344.84 Years.

This is assuming nicotine decays at an exponential rate. It could be linear, logistic, who knows.

This also assumes that whatever device this person used to measure nicotine was any sort of accurate and precise.

The point being, there is no hard science of the decay rate of nicotine in e-juice. Just try a google search, nothing but pure speculation.

I wish you good luck.

Thanks for agreeing with me.
 
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