Answer : The final temperature would be, 791.1 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 = 265 kJ/mol = 265000 J/mol
R = gas constant = 8.314 J/mole.K
= initial temperature = 
= final temperature = ?
Now put all the given values in this formula, we get:
![\log (\frac{4\times K_1}{K_1})=\frac{265000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{733K}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B4%5Ctimes%20K_1%7D%7BK_1%7D%29%3D%5Cfrac%7B265000J%2Fmol%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B733K%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
Therefore, the final temperature would be, 791.1 K
Answer: HYDROGEN BONDS
Explanation:
Water molecules attract each other happily thanks to their polarity. A hydrogen atom plus end associates an oxygen atom minus end.
These attractions are an example of hydrogen bonds, weak interactions forming between a partially positive charged hydrogen and a more electronegative atom like oxygen. The hydrogen atoms involved in bonding with hydrogen need to be bound to electronegative atoms such as Oxygen and fluorine
Answer: Your friend is incorrect.
Explanation: If we have an object or something that isn’t moving, (let’s say a notebook on a desk). If there is change, and the notebook moves, there is acceleration. Force = Mass times acceleration, f = m*a. There has to be a force, first of all. If you touched the notebook and moved it, some of your energy is transferred and now the notebook has kinetic energy. If our system is you and the notebook, the total energy doesn’t change. the energy is transferred, but doesn’t change. Your friend is not correct. Please give brainliest hope this helped!
Answer:
The value of the silver in the coin is 35.3 $
Explanation:
First of all, let's calculate the volume of the coin.
2π . r² . thickness = volume
r = diameter/2
r = 41 mm/2 = 20.5 mm
2 . π . (20.5 mm)² . 2.5 mm = 6601 mm³
Now, this is the volume of the coin, so we must find out how many grams are on it.
6601 mm³ / 1000 = 6.60 cm³
Let's apply density.
D = Mass / volume
10.5 g/cm³ = mass /6.60 cm³
10.5 g/cm³ . 6.60 cm³ = mass
69.3 g = mass
Each gram has a cost of 0.51$
69.3 g . 0.51$ = 35.3 $
