The answer should be 0.5 because 6/12 is 0.5. It gives u the key equation
(2.32g/cm³) x (1kg/1000g)x(1 000 000 cm³/1m³) = 2320 kg/m³
1 ml= 1 cm³
Answer:
For this assignment, I have picked three famous soccer players from around the world. I will go over their Birth, work ethic, and downfalls. I guess that most soccer players' downfalls are going to be their injuries.
The first soccer player I found was Ricardo Kakà. Ricardo Kakà was born to Simone dos Santos, his mother, and Bosco Izecson Pereira Leite, his father, on April 22, 1982. His work ethic mostly consists of "doing it". One of Ricardo Kakà's greatest downfalls is his injuries.
The second player I found was Cristiano Ronaldo. Cristiano Ronaldo was born to Maria Dolores dos Santos and José Dinis Aveiro. He was introduced to soccer by his dad because worked at a soccer field. Cristiano Ronaldo's work ethic consisted of something that Ricardo Kakà had as well. "Get it done". Cristiano Ronaldo's downfall was that his father drank too much.
I was not surprised to see that, but I thought it was going to be injuries as well.
The final player that I picked is Luis Suárez. Luis Suárez was born to Sandra Diaz and Rodolfo Suárez. He joined his local youth team that played soccer and he loved it. His work ethic is the same as the last soccer players. "Do it." His major downfall was that he was a bad-tempered teen and he got mad at stuff. That didn't help him when his parents divorced.
Overall all of the soccer players are different and very important. They have all gone through different situations and have gotten through them through soccer.
Explanation:
This is not 300 words, it is 244. You can it. Make sure to change it so it sounds like you wrote it.
Answer : The temperature will be, 392.462 K
Explanation :
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
=
= rate constant at 
= 
= activation energy for the reaction = 66.41 kJ/mole = 66410 J/mole
R = gas constant = 8.314 J/mole.K
= initial temperature = 293 K
= final temperature = ?
Now put all the given values in this formula, we get:
![\log (\frac{3K_1}{K_1})=\frac{66410J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{293K}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B3K_1%7D%7BK_1%7D%29%3D%5Cfrac%7B66410J%2Fmole%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B293K%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
Therefore, the temperature will be, 392.462 K
