Answer
is: activation energy of this reaction is 212,01975 kJ/mol.<span>
Arrhenius equation: ln(k</span>₁/k₂) =
Ea/R (1/T₂ - 1/T₁).<span>
k</span>₁
= 0,000643 1/s.<span>
k</span>₂
= 0,00828 1/s.
T₁ = 622 K.
T₂ = 666 K.
R = 8,3145 J/Kmol.
<span>
1/T</span>₁ =
1/622 K = 0,0016 1/K.<span>
1/T</span>₂ =
1/666 K = 0,0015 1/K.<span>
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol ·
(-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol </span>· (-0,0001 1/K).<span>
Ea = 212019,75 J/mol = 212,01975 kJ/mol.</span>
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
Answer:
35.5450 will be rounded to 35.55
Explanation:
=35.5450
if the last digit is less than 5 then it will be ignored
=35.545
when the dropping digit is 5 then the retaining digit will increse by a factor of 1
=35.55
i hope this will help you
She would be 150 on the moon because the moon has half the mass of earth