Answer:
8.15 L I2
Explanation:
<u>So, I have the balanced equation of</u>: 2CuBr2 +4KI → 2CuI + 4KBr + I2
<u>Now, we do</u>:
241.6g / 1 mol KI / 1 mol I2 / 22.4 L
--------------------------------------------------------- = 8.15 L I2
/ 166g KI / 4 mol KI / 1 mol I2
Hope this helps! ^u^
Answer:
a
The answer is shown on the first uploaded image
b
![K_{w} = 2.5*10^{-14} J/kgK](https://tex.z-dn.net/?f=K_%7Bw%7D%20%3D%202.5%2A10%5E%7B-14%7D%20J%2FkgK)
![pOH =6.80](https://tex.z-dn.net/?f=pOH%20%20%3D6.80)
![[OH^{-}] =1.6*10^{-7} M](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%20%20%3D1.6%2A10%5E%7B-7%7D%20M)
Explanation:
The explanation is shown on the second and third uploaded image
Answer: 628.15 K
Explanation:
Add 273.15 to any temperature in Celsius to get it in kelvin
p
K
a
=
5.4
pH
=
3.7
Explanation:
p
K
a
is simply
−
log
10
(
K
a
)
. It is a constant at any temperature and does not depend on the molarity of the solution.
p
K
a
=
−
log
10
(
K
a
)
=
−
log
10
(
4.0
×
10
−
6
)
=
5.4Explanation: