Sprouts McGee
New Member
I am about three week from harv and wondering if the cartridges will do any good
Cº2Cartridges will they work
- Thread starter Sprouts McGee
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Droopy Dog
New Member
You will just be wasting money.
DD
DD
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Sprouts McGee
New Member
If we assume the 12 grams is the weight of the CO2, the magic constant you seek is 22.4 liters per one mole of gas at STP. Since CO2 is 12 + 32= 44 gm for one mole 22.4 liters that means one cartridge contains
So 12/44 * 22.4 = 6.1 liters
so how many liters of gas is usually dispursed
(these cartridges were free so no cost)
into a grow room at any given time
So 12/44 * 22.4 = 6.1 liters
so how many liters of gas is usually dispursed
(these cartridges were free so no cost)
into a grow room at any given time
I dont really know about liters. Heres a formula I got from one of my grow books.
First calculate your grow room in cubic ft. length x width x height
co2 is naturally at 300-350 ppm. Optimally you want to be at 1500 ppm
So multiply your room volume by 0.0012 to give you the amount needed in cubic ft. I dont know if that helps but its what I go by.
First calculate your grow room in cubic ft. length x width x height
co2 is naturally at 300-350 ppm. Optimally you want to be at 1500 ppm
So multiply your room volume by 0.0012 to give you the amount needed in cubic ft. I dont know if that helps but its what I go by.
use them and get more if you can. dont vent. let them suck it up. but a propane torch works better. i have a meter and get it up to 1800ppms w/ propane those co2 cartridges contain very small amounts of co2. i used to use them . the little silver bullets? for blowing out pipes lines. im an hvac tech. have a bunch of them.
its better than nothing. but your better off breathing on the plants.
its better than nothing. but your better off breathing on the plants.
Every mol of any gas expands to 22,4L gas (ideal gas law)
1 mol CO2 consists of 1 C (12g/mol) and 2 O (16g/mol). So CO2 has: 12+16+16=44g/mol
44g of CO2 will subsequently have 22,4L of volume
If we have 12g of CO2 it means:
44g=100% 12g=27,3%
100%=22,4L 27,3%=(22,4*0,273)=6,1L
Conclusion: A cartridge containing 12g of CO2 will expand to aprox. 6,1L of gas (aprox. 2 1/2 gal)
Its called "stoichiometry" and its what I do every day for a living ^^
PS:
Ideal gas law: 1mol(gas)=22,4l in 0°C and 1013 mbar
1mol=6,022*10^23 atoms/molecules
or:
1mol=as many atoms as in 12g C^12
Mol is not really a "weight" but a solid number of atoms/molecules. It means an atom/molecule with the mol weight of 12 (like carbon) will weigh 12g/mol.
This concept applies to all chemical reaction equations:
1 mol H2+1/2 mol O2->(T+)->1 mol H20 (or 2 H2 + O2 = 2 H2O)
H2=2g/mol
O=16g/mol or O2=32g/mol
(T+)=heat
so:
If you mix 2g of Hydrogen gas with 16g of Oxigen gas and then light it up, it will go BOOOM and leave you with 18g of water.
1 mol CO2 consists of 1 C (12g/mol) and 2 O (16g/mol). So CO2 has: 12+16+16=44g/mol
44g of CO2 will subsequently have 22,4L of volume
If we have 12g of CO2 it means:
44g=100% 12g=27,3%
100%=22,4L 27,3%=(22,4*0,273)=6,1L
Conclusion: A cartridge containing 12g of CO2 will expand to aprox. 6,1L of gas (aprox. 2 1/2 gal)
Its called "stoichiometry" and its what I do every day for a living ^^
PS:
Ideal gas law: 1mol(gas)=22,4l in 0°C and 1013 mbar
1mol=6,022*10^23 atoms/molecules
or:
1mol=as many atoms as in 12g C^12
Mol is not really a "weight" but a solid number of atoms/molecules. It means an atom/molecule with the mol weight of 12 (like carbon) will weigh 12g/mol.
This concept applies to all chemical reaction equations:
1 mol H2+1/2 mol O2->(T+)->1 mol H20 (or 2 H2 + O2 = 2 H2O)
H2=2g/mol
O=16g/mol or O2=32g/mol
(T+)=heat
so:
If you mix 2g of Hydrogen gas with 16g of Oxigen gas and then light it up, it will go BOOOM and leave you with 18g of water.
well that sure clears things up.......
Ohhh stoichiometry can be so incredibly usefull. I need it like every day. One thing that might be of special interest for Growers ist the:
"Common Mixture Equasion"
(Dont be afraid of this, its very very simple!)
m(t)*c(t)=m(1)*(c1)+m(2)*c(2)+....m(f)*c(f)
With:
m(t)=total/final mass or volume
c(t)=total/final concentration
m(1)=mass/volume of ingredient 1
c(1)=concentration of ingredient 1
m(2)=....... of ingredient 2
c(2)=....... of ingredient 2
and so on...
Lets say your tap water has an exact EC of 0,78 and you want to mix it with destilled water (EC=0) until you have an exact EC of 0,2. First you decide how much tap water you want to use; to make it simple we will take one liter. We will set the ammount of destilled water needed as X Liters and the tap water is as said 1L; so m(water total)=1L+XL
Lets put the information we have in the equation:
1L+XL(Water total)*c(0,2)=1L(tap water)*c(0,78)+XL(destilled water)*c(0)
Lets make this a proper equation:
(1+X)*(0,2)=1*0,78+0*X
Now we solve to X:
0,2+0,2X=0,78+0 :
0,2+0,2X=0,78 : -0,2
0,2X=0,58 : /0,2X
x=2,9
Conclusion:
If you mix 1 Liter of tapwater with an EC of 0,78 with 2,9 Liters of destilled water with an EC of 0 than you will get 3,9 Liters of water with an EC of 0,2.
Checking:
Lets say the tap water has an EC of 0,8 and the destilled 0. So
1L tw + 1L dw = 2L EC 0,4
2L EC 0,4 + 2L EC 0 = 4L EC 0,2
---> The original equation/answer is most likely to be correct!!!
Easy as that! Works everytime and makes life in a laboratoty MUCH easier.
Have fun mixing!
"Common Mixture Equasion"
(Dont be afraid of this, its very very simple!)
m(t)*c(t)=m(1)*(c1)+m(2)*c(2)+....m(f)*c(f)
With:
m(t)=total/final mass or volume
c(t)=total/final concentration
m(1)=mass/volume of ingredient 1
c(1)=concentration of ingredient 1
m(2)=....... of ingredient 2
c(2)=....... of ingredient 2
and so on...
Lets say your tap water has an exact EC of 0,78 and you want to mix it with destilled water (EC=0) until you have an exact EC of 0,2. First you decide how much tap water you want to use; to make it simple we will take one liter. We will set the ammount of destilled water needed as X Liters and the tap water is as said 1L; so m(water total)=1L+XL
Lets put the information we have in the equation:
1L+XL(Water total)*c(0,2)=1L(tap water)*c(0,78)+XL(destilled water)*c(0)
Lets make this a proper equation:
(1+X)*(0,2)=1*0,78+0*X
Now we solve to X:
0,2+0,2X=0,78+0 :
0,2+0,2X=0,78 : -0,2
0,2X=0,58 : /0,2X
x=2,9
Conclusion:
If you mix 1 Liter of tapwater with an EC of 0,78 with 2,9 Liters of destilled water with an EC of 0 than you will get 3,9 Liters of water with an EC of 0,2.
Checking:
Lets say the tap water has an EC of 0,8 and the destilled 0. So
1L tw + 1L dw = 2L EC 0,4
2L EC 0,4 + 2L EC 0 = 4L EC 0,2
---> The original equation/answer is most likely to be correct!!!
Easy as that! Works everytime and makes life in a laboratoty MUCH easier.
Have fun mixing!
A more complex equation:
Mix water with EC=0,78 with badly destilled water EC=0,1 until you got water with an EC=0,54!
Again we will use 1L of water with XL of destilled water.
(1+X)*(0,54)=1*0,78+X*0,1
0,54+0,54X=0,78+0,1X : -0,54
0,54X=0,24+0,1X : -0,1X
0,44X=0,24 : /0,44
X=0,55
So, mix 1L of tap water with 0,55L of the badly destilled water to get 1,55L with an EC of 0,54.
Its really fast and easy.
NOTE; This only works with linear concentrations. You CAN NOT use it to calculate pH mixtures as they are not linear but exponential!!! But works fine with fertilisers etc.
Mix water with EC=0,78 with badly destilled water EC=0,1 until you got water with an EC=0,54!
Again we will use 1L of water with XL of destilled water.
(1+X)*(0,54)=1*0,78+X*0,1
0,54+0,54X=0,78+0,1X : -0,54
0,54X=0,24+0,1X : -0,1X
0,44X=0,24 : /0,44
X=0,55
So, mix 1L of tap water with 0,55L of the badly destilled water to get 1,55L with an EC of 0,54.
Its really fast and easy.
NOTE; This only works with linear concentrations. You CAN NOT use it to calculate pH mixtures as they are not linear but exponential!!! But works fine with fertilisers etc.
And the most complex and very last equasion:
You got 43,8L of soil, containing 12% worm manure. Mix it with pure worm manure until it has 28% of it. Sadly you only have 1,2L pure worm manure left but also some other soil containing almost 87% of the worm castings. How much do you need of the other soil if you want to use up your pure manure first.
(43,8+1+X)*(28)=43,8*12+1,2*100+X*87
(44,8+X)*(28)=525,6+120+87X
1254,4+28X=645,6+87X
28X=-608,8+87X
-59X=-608,8
X=10,32
You will need to mix in 10,32L of the second soil together with the pure worm manure to get the desired soil with 28% worm manure!
You see, this really can make the (seemingly) most uncomfortable mixing problems very easy.
SORRY FOR THE COMPLETE OT ^^
You got 43,8L of soil, containing 12% worm manure. Mix it with pure worm manure until it has 28% of it. Sadly you only have 1,2L pure worm manure left but also some other soil containing almost 87% of the worm castings. How much do you need of the other soil if you want to use up your pure manure first.
(43,8+1+X)*(28)=43,8*12+1,2*100+X*87
(44,8+X)*(28)=525,6+120+87X
1254,4+28X=645,6+87X
28X=-608,8+87X
-59X=-608,8
X=10,32
You will need to mix in 10,32L of the second soil together with the pure worm manure to get the desired soil with 28% worm manure!
You see, this really can make the (seemingly) most uncomfortable mixing problems very easy.
SORRY FOR THE COMPLETE OT ^^
:surrender:
ChicagoJoe
420 Member
my sentiments exactly.
Panic
New Member
Dude if you hadn't lost me at "stoichiometry" you surely would have lost me with: m(t)*c(t)=wtf(1)*c1)+more wtf*c(2)......
