Answer:
The activation energy is 7.11 × 10⁴ J/mol.
Explanation:
Let's consider the Arrhenius equation.

where,
k is the rate constant
A is a collision factor
Ea is the activation energy
R is the ideal gas constant
T is the absolute temperature
The plot of ln k vs 1/T is a straight line with lnA as intercept and -Ea/R as slope. Then,

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
Isotope 1: 89.905 * 51.45 = 4625.61225 / 100 = 46.2561225
Isotope 2: 90.906 * 11.22 = 1019.96532 / 100 = 10.1996532
Isotope 3: 91.905 * 17.15 = 1576.17175 / 100 = 15.7617075
Isotope 4: 93.906 * 17.38 = 1632.08628 / 100 = 16.3208628
Isotope 5: 95.908 * 2.08 = 268.5424 / 100 = 2.685424
46.2561225 + 10.1996532 + 15.7617075 + 16.3208628 + 2.685424 = 91.22377
actual mass Zr = about 91.22
Answer:
heat energy because u are boiling something