PRICE IN
FIRMS
"B"
MARKET SUPPLY
(K.
A
"C"
10
10
50
80
20
40
45
25
3:
4.
30
40
50
120
to Complete the quantity of onion supplied by
forms to the market uh above table
Draw the Market supply oww
schedule and explain ut
from the
O Option 1
Answers
Answer:
Scientific progress and the COVID-19 pandemic
Most recent answer
8th Aug, 2020
Mahdy H. Hamed
New Valley University
Two kg soil was need to 24 gm for N, 12 gm for K, and 10 gm for P
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Popular Answers (1)
Deleted profile
Hi Mohammed
Usually fertilizers sold with a formula N-P-K are expressed in N- P2O5-and K2O. You can see this from the fertilizers you want to use.
So now I guess that if you said you want N-P-K at 120-60-50 kg/ha rate that these rates are really for N-P-K and not N- P2O5 and K2O.
If not the conversion factors are:
P2O5 x 0.4364 = P
P x 2.20 = P2O5
K2O x 0.83 = K
K x 1.20 = K2O
These conversion factors are based on the atomic mass:
O = 15.99
K = 39.0983
P = 30.973
Ex.
K2O= = (2 x 39.0983) + 15.99 = 78.1966 + 15.99 = 94.1866
So there is 2 K in K2O
There is 78.1966 (K) in 94.1866 (K2O_
There is 1 K in x
X = 78.1966 / 94.1866 = 0.83
So the conversion factor is K2O x 0.83 = K
The other way around is 94.1866 / 78.1966 = 1.204 thus the conversion factor is
1.2 x K = K2O
You can do the same exercise for P2O5
Now there is two way to apply your fertilizers either by calculating the surface of 1 ha and the surface of your pots, or you calculate the weight of one ha and we know that you have 2 kg in each pot.
Now the weight of 1 ha of soil will greatly depend on the soil bulk density, the nature of soil (minreal or organic, its organic matter content ect…).
Since we don’t have the information about your soil we can consider that it is a typical mineral soil and its hectare –furrow slice of its surface soil will weight 2.2 million kilograms.
To simplify your calculation you can consider 2 million kilograms.
Now the determination of your need of N-P-K at 120-60-50 kg/ha rate I guess was performed on the basis of soil analysis and the crop you want to grow. Now these needs are for the crop until harvest, will you be attaining this stage in pots? Let’s us assume that you will.
Lets us calculate for N
You need are
120 Kg N in 2000000 Kg of soil
120 000g N in 2000000 Kg or 120 g in 2000 Kg of soil
120 000 N mg in 2000 Kg of soil
Or 120 mg N in 2 Kg of soil
Now you ammonium nitrate contains 33 % of N
So 100 mg of ammonium nitrate contains 33 mg of N
X mg of ammonium nitrate contains 120 mg of N
X = (120 X 100) / 33 = 363.636 mg ammonium nitrate have to be added to 2 Kg of soil
Now for practical reason you will not mixt this quantity in the soil of each pot, you should prepare the mixture of soil and fertilizer to all pots, by mixing first the fertilizer to a small quantity of soil, and increasing the soil quantity gradually to assure a uniform distribution of your fertilizer in the total amount of soil you are using in your experiment. This remark also apply for other fertilzers
Now let me do the P2O5 for you and then you can calculate the K2O yourself easily.
So you need
60 Kg of P per 2 000 000 Kg of soil
60 000 000 mg of P per 2000 000 Kg of soil
60 mg of P per 2 Kg of soil
Your fertilizer is
Triple super phosphate 46 % P2O5
So 100 Kg of triple superphosphate contain 46 Kg of P2O5
considering the conversion factor
P2O5 x 0.4364 = P
So 100 kg of triple superphosphate contain 20.07 Kg of P
100 mg of triple superphosphate contain 20 mg of P
300 mg of triple superphosphate contain 60 mg of P
Add to each pot (2 kg of soil) 300 mg of triple superphosphate.
And see the remark for N you need to prepare the mixture for all the soil requested for your experiment .
Can you now do the calculation for K? If not send me a mail.
It will be nice if somebody double check my calculations
Now if your experiment will last for a long time you might want to consider fractionation of N because you might lose it in the leachates while watering plants..
Explanation:
plz thank me