Somethings that are beneficial is its good for plants, if it is very fine ash it is able to break down and quickly dissolve into the soil.
Explanation:
Given that,
The mass of the object, m = 5 kg
We need to find the maximum potential energy. When the height of the object is maximum it will have the maximum potential energy.
The attached figure shows that the maximum height is 25 m. The formula for the potential energy is given by :
![P=mgh\\\\P=5\times 9.8\times 25\\\\P=1225\ J](https://tex.z-dn.net/?f=P%3Dmgh%5C%5C%5C%5CP%3D5%5Ctimes%209.8%5Ctimes%2025%5C%5C%5C%5CP%3D1225%5C%20J)
So, when the height is 25 m, the object will have the highest potential energy.
Answer:
![k_2=2.46](https://tex.z-dn.net/?f=k_2%3D2.46)
Explanation:
Hello!
In this case, according to the Arrhenius equation for variable temperature:
![ln(\frac{k_2}{k_1} )=-\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )](https://tex.z-dn.net/?f=ln%28%5Cfrac%7Bk_2%7D%7Bk_1%7D%20%29%3D-%5Cfrac%7BEa%7D%7BR%7D%28%5Cfrac%7B1%7D%7BT_2%7D%20-%5Cfrac%7B1%7D%7BT_1%7D%20%29)
Given the temperatures to be computed in kelvins and the activation energy, we obtain:
![ln(\frac{k_2}{0.924} )=-\frac{89800J/mol}{8.314J/(mol*K)}(\frac{1}{306.45K} -\frac{1}{298.15K} )\\\\ln(\frac{k_2}{0.924} )=0.981\\\\k_2=0.924*exp(0.981)\\\\k_2=2.46](https://tex.z-dn.net/?f=ln%28%5Cfrac%7Bk_2%7D%7B0.924%7D%20%29%3D-%5Cfrac%7B89800J%2Fmol%7D%7B8.314J%2F%28mol%2AK%29%7D%28%5Cfrac%7B1%7D%7B306.45K%7D%20-%5Cfrac%7B1%7D%7B298.15K%7D%20%29%5C%5C%5C%5Cln%28%5Cfrac%7Bk_2%7D%7B0.924%7D%20%29%3D0.981%5C%5C%5C%5Ck_2%3D0.924%2Aexp%280.981%29%5C%5C%5C%5Ck_2%3D2.46)
Best regards!
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2, valence shell: 4s2 3d10 4p2
This is a Charles' Law problem: V1/T1 = V2/T2. As the temperature of a fixed mass of gas decreases at a constant pressure, the volume of the gas should also decrease proportionally. Here, we are given the new volume of the gas after cooling, and we want to determine to what temperature the gas was cooled.
To use Charles' Law, the temperature must be in Kelvin (x °C = x + 273.15 K). We want to solve Charles' Law for T2, which we can obtain by rearranging the equation into T2 = V2T1/V1. Given V1 = 130 L, T1 = 250 °C (523.15 K), and V2 = 85 L:
T2 = (85 L)(523.15 K)/(130 L) = 342 K or 68.9 °C. If sig figs are to be considered, since all the values in the question are given to two sig figs, the answer to two sig figs would be either 340 K or 69 °C.