Answer:
To the best of my knowledge, it is because of the amount of gamma rays is given off.
Explanation:
While both are isotopes, Potassium 40 gives off fewer gamma rays compared to Cobalt 60. Potassium 40 isn't really harmful to humans, but Cobalt 60 (I believe) is used in chemotherapy.
Answer:
a) ![\triangle G^{0} = 7.31 kJ/mol](https://tex.z-dn.net/?f=%5Ctriangle%20G%5E%7B0%7D%20%3D%207.31%20kJ%2Fmol)
b) ![K_{-1} = 0.0594 m^{-1} s^{-1}](https://tex.z-dn.net/?f=K_%7B-1%7D%20%3D%200.0594%20m%5E%7B-1%7D%20s%5E%7B-1%7D)
Explanation:
Equation of reaction:
![2 XO_{2} (g) + O_{2} (g) \rightleftharpoons 2XO_{3} (g)](https://tex.z-dn.net/?f=2%20XO_%7B2%7D%20%28g%29%20%2B%20O_%7B2%7D%20%28g%29%20%5Crightleftharpoons%202XO_%7B3%7D%20%28g%29)
Initial pressure 3 1 0
Pressure change 2P 1P 2P
Total pressure = (3-2P) + (1-P) + (2P)
Total Pressure = 3.75 atm
(3-2P) + (1-P) + (2P) = 3.75
4 - P = 3.75
P = 4 - 3.75
P = 0.25 atm
Let us calculate the pressure of each of the components of the reaction:
Pressure of XO2 = 3 - 2P = 3 - 2(0.25)
Pressure of XO2 =2.5 atm
Pressure of O2 = 1 - P = 1 -0.25
Pressure of O2 = 0.75 atm
Pressure of XO3 = 2P = 2 * 0.25
Pressure of XO3 = 0.5 atm
From the reaction, equilibrium constant can be calculated using the formula:
![K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20%5Cfrac%7B%5BPXO_%7B3%7D%5D%20%5E%7B2%7D%20%7D%7B%5BPXO_%7B2%7D%5D%20%5E%7B2%7D%5BPO_%7B2%7D%5D%20%7D)
![K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20%5Cfrac%7B0.5%5E2%7D%7B2.5%5E2%20%2A0.75%7D%20%5C%5CK_%7Bp%7D%20%3D%200.0533%20%3D%20K_%7Beq%7D)
Standard free energy:
![\triangle G^{0} = - RT ln k_{eq} \\\triangle G^{0} = -(0.008314*300* ln0.0533)\\\triangle G^{0} = 7.31 kJ/mol](https://tex.z-dn.net/?f=%5Ctriangle%20G%5E%7B0%7D%20%3D%20-%20RT%20ln%20k_%7Beq%7D%20%5C%5C%5Ctriangle%20G%5E%7B0%7D%20%3D%20-%280.008314%2A300%2A%20ln0.0533%29%5C%5C%5Ctriangle%20G%5E%7B0%7D%20%3D%207.31%20kJ%2Fmol)
b) value of k−1 at 27 °C, i.e. 300K
![K_{1} = 7.8 * 10^{-2} m^{-2} s^{-1}](https://tex.z-dn.net/?f=K_%7B1%7D%20%3D%207.8%20%2A%2010%5E%7B-2%7D%20m%5E%7B-2%7D%20s%5E%7B-1%7D)
![K_{c} = K_{p}RT\\K_{c} = 0.0533* 0.0821 * 300\\K_{c} = 1.313 m^{-1}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20K_%7Bp%7DRT%5C%5CK_%7Bc%7D%20%3D%200.0533%2A%200.0821%20%2A%20300%5C%5CK_%7Bc%7D%20%3D%201.313%20m%5E%7B-1%7D)
![K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2} }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}](https://tex.z-dn.net/?f=K_%7B-1%7D%20%3D%20%5Cfrac%7BK_%7B1%7D%20%7D%7BK_%7Bc%7D%20%7D%20%5C%5CK_%7B-1%7D%20%3D%20%5Cfrac%7B7.8%20%2A%2010%5E%7B-2%7D%20%20%7D%7B1.313%20%7D%5C%5CK_%7B-1%7D%20%3D%200.0594%20m%5E%7B-1%7D%20s%5E%7B-1%7D)
Answer:
a scientific question
Procedure
Conclusion.
Explanation:
Scientific investigation is usually carried out to explain and give answers to natural phenomenon. Now, this scientific investigation is performed through the three main types as described in the question.
With that being said, the most common parts of all types of investigation are :
1. Scientific question: In this stage, questions are usually asked on the basis of observation made which could be answered.
2. Procedure: In this stage, the dimension of the steps are followed to answer and test the scientific question.
3. Conclusion: This is the primary aim that we check when done with our sample values