Answer:
it will dilute to its natrual state. so c
Explanation:
Answer:
m = 0.659 ounce
Explanation:
It is given that,
The thickness of a Teflon coating is, d = 1 mm
Area of the coating, A = 36 inch²
The density of Teflon, d = 0.805 g/mL
We need to find ounces of Teflon are needed.
Firstly, find the volume of the Teflon needed,
1 inch² = 6.4516 cm²
36 inch² = 232.258 cm²
Density,

V is volume of the Teflon needed, V = Ad
So,

Also, 1 gram = 0.035274 ounce
18.69 gram = 0.659 ounce
So, 0.659 ounces of Teflon are needed.
<u>Given:</u>
Initial temperature, T1 = 20 C
Final temperature, T2 = 40 C
<u>To determine:</u>
The temperature change
<u>Explanation:</u>
Convert degree C to Kelvin
Temperature in Kelvin = degree C + 273
T1 = 20 + 273 = 293 K
T2 = 40 + 273 = 313 K
Incremental temperature change = T2 - T1 = 313-293 = 20 K
Ans: The temperature change in kelvin is 20 K
Answer: The rate increases 3 times on raising the temperature from 20degree to 30 degree
Explanation:
According to Arrhenius equation with change in temperature, the formula is as follows.
![ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]](https://tex.z-dn.net/?f=ln%20%5Cfrac%7Bk_%7B2%7D%7D%7Bk_%7B1%7D%7D%20%3D%20%5Cfrac%7B-E_%7Ba%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D%20-%20%5Cfrac%7B1%7D%7BT_%7B1%7D%7D%5D)
where
= rate constant at temp 
= rate constant at temp
= activation energy
R= gas constant
= temperature = 
= temperature = 
![ln \frac{k_{2}}{k_{1}} = \frac{-85\times 1000J/mol}{8.314J/Kmol}[\frac{1}{303} - \frac{1}{293}]](https://tex.z-dn.net/?f=ln%20%5Cfrac%7Bk_%7B2%7D%7D%7Bk_%7B1%7D%7D%20%3D%20%5Cfrac%7B-85%5Ctimes%201000J%2Fmol%7D%7B8.314J%2FKmol%7D%5B%5Cfrac%7B1%7D%7B303%7D%20-%20%5Cfrac%7B1%7D%7B293%7D%5D)


Thus rate increases 3 times on raising the temperature from 20degree to 30 degree