<u>Answer:</u> The partial pressure of gas number 3 is 159 mmHg
<u>Explanation:</u>
Dalton's law of partial pressure states that the total pressure of the system is equal to the sum of partial pressure of each component present in it.
To calculate the partial pressure of gas number 3, we use the law given by Dalton, which is:

We are given:
Total pressure of the tank,
= 950 mmHg
Vapor pressure of gas number 1,
= 335 mmHg
Vapor pressure of gas number 2,
= 456 mmHg
Putting values in above equation, we get:
![950=335+456+p_3\\\\p_{3}=[950-(335+456)]=159mmHg](https://tex.z-dn.net/?f=950%3D335%2B456%2Bp_3%5C%5C%5C%5Cp_%7B3%7D%3D%5B950-%28335%2B456%29%5D%3D159mmHg)
Hence, the partial pressure of gas number 3 is 159 mmHg
Answer:
49°C
Explanation:
Let's apply the Ideal Gases Law in order to solve this question:
P . V = n . R . T
Pressure = 1 atm
Volume = 6 L
n = number of moles → 10 g. 1mol /44g = 0.227 moles
R = Ideal Gases Constant
We replace data: 1 atm . 6 L = 0.227 mol . 0.082 . T
6 atm.L / ( 0.227 mol . 0.082) = T
T° = 322 K
We convert T° from K to °C → 322 K - 273 = 49°C
Answer:
36.23 %
Explanation:
Let's <em>assume we have 1 mol of FeCl₂ · 4H₂O</em>. In that case we would have:
- 1 mol of FeCl₂, weighing 126.75 g (that's the molar mass of FeCl₂), and
- 4 moles of H₂O, weighing (4 * 18 g/mol) 72 g.
Now we can <u>calculate the percent by mass of water</u>:
- % mass = mass of water / total mass * 100%
- % mass =
* 100% = 36.23 %
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