The definition of heat transfer through convection involves the movement of a fluid that causes heat to move away from a hear source. That being said, convection can take place in a lake, air inside, and air outside since both liquids and gasses are considered to be fluids.
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Gold is actually very soft. If rings were made of pure gold they would bend and become disfigured. Other elements are added to give the ring its stability.
Since the reaction shown in the question is an acid - base reaction in the Lewis sense; the Lewis acid here is AlCl3 while the Lewis base here is Cl^- .
<h3>What is a Lewis acid?</h3>
A Lewis acid is a substance that accepts electron pair while a Lewis base donates an electron pair.
Now consider the given reaction; AlCl3 +Cl^- ------> AlCl 4 ^-. The Lewis acid here is AlCl3 while the Lewis base here is Cl^- .
Learn more about acid - base reaction: brainly.com/question/14356798
Answer:
0.234 M
Explanation:
C- 12.009 x 7
H- 1.001 x 5
N- 14.006
O- 16 x 3
S- 32.059
___________+
183.133 g/mol
= 0.234 M Cancel out the grams mol/L equals molarity. Lowest significant figure is 3
Answer: ![4.3\times 10^{-13}s^{-1}](https://tex.z-dn.net/?f=4.3%5Ctimes%2010%5E%7B-13%7Ds%5E%7B-1%7D)
Explanation:
According to the Arrhenius equation,
![K=A\times e^{\frac{-Ea}{RT}}](https://tex.z-dn.net/?f=K%3DA%5Ctimes%20e%5E%7B%5Cfrac%7B-Ea%7D%7BRT%7D%7D)
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
= ![6.1\times 10^{-8}s^{-1}](https://tex.z-dn.net/?f=6.1%5Ctimes%2010%5E%7B-8%7Ds%5E%7B-1%7D)
= rate constant at
= ![?](https://tex.z-dn.net/?f=%3F)
= activation energy for the reaction = 262 kJ/mol = 262000J/mol
R = gas constant = 8.314 J/mole.K
= initial temperature = ![600.0K](https://tex.z-dn.net/?f=600.0K)
= final temperature = ![775.0K](https://tex.z-dn.net/?f=775.0K)
Now put all the given values in this formula, we get
![\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B6.1%5Ctimes%2010%5E%7B-8%7D%7D%7BK_2%7D%29%3D%5Cfrac%7B262000%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B600.0K%7D-%5Cfrac%7B1%7D%7B775.0K%7D%5D)
![\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B6.1%5Ctimes%2010%5E%7B-8%7Ds%5E%7D%7BK_2%7D%29%3D5.150)
![(\frac{6.1\times 10^{-8}}{K_2})=141253.8](https://tex.z-dn.net/?f=%28%5Cfrac%7B6.1%5Ctimes%2010%5E%7B-8%7D%7D%7BK_2%7D%29%3D141253.8)
Therefore, the value of the rate constant at 775.0 K is ![4.3\times 10^{-13}s^{-1}](https://tex.z-dn.net/?f=4.3%5Ctimes%2010%5E%7B-13%7Ds%5E%7B-1%7D)