well, let me see if I can tackle this for you.
Asking how much K in KNO3 is tricky. It all depends on how you define the question. it could be 1) how many grams of K for every gram of KNO3. it could also mean how many moles of K per gram of NO3 etc etc.
lets consider the 1st more obvious possiblility. the mass of on mole of the following (rounded for easier math)
K=39g
N=14g
O=16g
so 1 mole of KNO3=39+14+16=69g. so 39/69=~57% K.
You can take the above example and apply it so many other ways using different definitions of "how much"
To answer you second part is very simple. It comes down to the following formula
M1V1=M2V2 where V=volumes in liters and M =concentrations in moles/L
alternately, it could be grams and milliliters...all depends on how things are defined in the starting values
Example. Suppose Kent is 10% Ca (10^-2 ...ppm=10^-6 so 10% Ca=10000ppm)
You want to add 1mL of 10,000 ppm Ca into your 29g tank.
we need to 1st convert 25g to L which well approximate at ~100L
M1V1=M2V2
(10,000ppm Ca)*(1ml *1L/1000mL)=M2*100L
Solve for M2 and you find out that you just added 0.1ppm Ca
(NOTE, I have rounded a lot of stuff for simplicity)
The 104% deal is kinda tricky and depends on how they define their units...which we can't tell right now.
As for the fluffy thing, recall that density has the unitys of mass/volume. as long as you know the approximate density of the material and account for it, you can measure out stuff approximately by using volumes (tsp/tbs) instead of mass (grams)
Hope this clears some stuff up
...hope I didn't make a stupid mistake in my rapid calculations too
Asking how much K in KNO3 is tricky. It all depends on how you define the question. it could be 1) how many grams of K for every gram of KNO3. it could also mean how many moles of K per gram of NO3 etc etc.
lets consider the 1st more obvious possiblility. the mass of on mole of the following (rounded for easier math)
K=39g
N=14g
O=16g
so 1 mole of KNO3=39+14+16=69g. so 39/69=~57% K.
You can take the above example and apply it so many other ways using different definitions of "how much"
To answer you second part is very simple. It comes down to the following formula
M1V1=M2V2 where V=volumes in liters and M =concentrations in moles/L
alternately, it could be grams and milliliters...all depends on how things are defined in the starting values
Example. Suppose Kent is 10% Ca (10^-2 ...ppm=10^-6 so 10% Ca=10000ppm)
You want to add 1mL of 10,000 ppm Ca into your 29g tank.
we need to 1st convert 25g to L which well approximate at ~100L
M1V1=M2V2
(10,000ppm Ca)*(1ml *1L/1000mL)=M2*100L
Solve for M2 and you find out that you just added 0.1ppm Ca
(NOTE, I have rounded a lot of stuff for simplicity)
The 104% deal is kinda tricky and depends on how they define their units...which we can't tell right now.
As for the fluffy thing, recall that density has the unitys of mass/volume. as long as you know the approximate density of the material and account for it, you can measure out stuff approximately by using volumes (tsp/tbs) instead of mass (grams)
Hope this clears some stuff up
...hope I didn't make a stupid mistake in my rapid calculations too
