Answer:
a. 300 kg of Fertilizer
b. 225 kg of fertilizer
c.400 Kg of fertilizer
d.600 Kg of fertilizer
Explanation:
The percentage composition ratio of Nitrogen, Phosphorus and Potassium bag of the given fertilizer is 40:15:10.
The percentages can be expressed as fractions as follows:
For nitrogen; 40/100 = 0.4
For phosphorus; 15/100 = 0.15
For potassium; 10/100 = 0.1
To find the quantity of fertilizer required to add to a hectare to supply the given amount of nutrients, the amount to be provided is divided by the percentage or fractional compostion of each nutrient.
Quantity of fertilizer required to add to a hectare to supply;
a. Nitrogen at 120 kg/ha = 120/0.4 = 300 Kg of fertilizer
b.. Nitrogen at 90 Kg/ha = 90/0.4 = 225 Kg of fertilizer
c. Phosphorus at 60 kg/ha = 60/0.15 = 400 Kg of fertilizer
d. Potassium at 60 kg/ha = 60/0.1 = 600 Kg of fertilizer
The sewage dumps were toxic so it would effect the air and cause many help issues especially with lungs.
Answer: Correct statements are given below:
1) In activity series elements are listed according to their reactivity.
2) Nuclear reaction in which low atomic nuclei fuse to form a heavier nucleus with the release of energy. Nuclear fusion reaction occurs in the sun.
3) C3H8 + 5O2 → 3CO2 + 4H2O is the example of a combustion reaction.
4) 2Fe + O2 → 2FeO is the example of synthesis reaction. In synthesis reaction is a type of reaction in which two reactants combine to form a single product.
Br₂ (l) + 2 NaI (s) → 2 NaBr (s) + I₂ (s)
Explanation:
Reacting bromide (Br₂) with sodium iodine (NaI) will produce sodium bromide (NaBr) and iodine (I₂).
To balance the equation the number of atoms of each element entering the reaction have to be equal to the number of atoms of each element leaving the reaction, in order to conserve the mass.
Br₂ (l) + 2 NaI (s) → 2 NaBr (s) + I₂ (s)
where:
l - liquid
s - solid
This is a single replacement reaction because an element in a compound is replaced by another element. Generally a single replacement reaction is represented as: A + BC → AC + B
Learn more about:
types of chemical reactions
brainly.com/question/10105284
balancing chemical equations
brainly.com/question/13908054
#learnwithBrainly
Answer: 1. 
moles
2. 90 mg
Explanation:
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus 
moles of ozone is removed by =
moles of sodium iodide.
Thus moles of sodium iodide are needed to remove 
moles of 
2. 
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus 0.0003 moles of ozone is removed by =
moles of sodium iodide.
Mass of sodium iodide= 
(1g=1000mg)
Thus 90 mg of sodium iodide are needed to remove 13.31 mg of .
