Answer:
260K
Explanation:
We use the general gas equation to solve this. The mathematical form can be represented as follows:
P1V1/T1 = P2V2/T2
We are to look for T2 from the question, rearranging the equation yields:
T2 = T1P2V2/P1V1
So what information do we have from the question:
T2= ?
P2 = 640mmHg
V2 = 198L
P1 = 760mmHg
T1 = 273K as they gas is at S.T.P since pressure is 760mmHg which is the standard pressure
V1 = 175L
We now substitute all these values:
T2 = (273 * 640 * 198)/(760 * 175)
T2 = 260K
<u>Answer:</u> The 
of the acid is 6.09
<u>Explanation:</u>
For the given chemical reaction:
The expression of equilibrium constant [tex[(K_a)[/tex] for the above equation follows:
![K_a=\frac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
We are given:
![[HA]_{eq}=0.200M](https://tex.z-dn.net/?f=%5BHA%5D_%7Beq%7D%3D0.200M)
![[H^+]_{eq}=4.00\times 10^{-4}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D_%7Beq%7D%3D4.00%5Ctimes%2010%5E%7B-4%7DM)
![[A^-]_{eq}=4.00\times 10^{-4}M](https://tex.z-dn.net/?f=%5BA%5E-%5D_%7Beq%7D%3D4.00%5Ctimes%2010%5E%7B-4%7DM)
Putting values in above expression, we get:
p-function is defined as the negative logarithm of any concentration.
So,
Hence, the of the acid is 6.09
Answer: 98.08 g/mol
Explanation:
i just took this test and got it right
Answer:
Dust and smoke.
Explanation:
Dust and smoke are two different particles present in the air. Dust and smoke are different from one another due to their origin. Smoke formed from burning of materials while dust refers to the soil particles lifted by the wind due to their light weight. Dust and smoke are similar to each other due to their small in size, infinite number means uncountable and light weight.
