The sample of oxygen gas was collected through water displacement. So, the gas collected will be a mixture of oxygen and water vapor.
Given that the total pressure of the mixture of gases containing oxygen and water vapor = 749 Torr
Vapor pressure of pure water at 
=25.81mmHg
= 
According to Dalton's law of partial pressures,
Total pressure = Partial pressure of Oxygen gas + Partial pressure of water
749 Torr = Partial pressure of Oxygen gas + 25.81 Torr
Partial pressure of Oxygen gas = 749 Torr - 25.81 Torr = 723.19 Torr
Therefore the partial pressure of Oxygen gas in the mixture collected will be 723.19 Torr
Answer:
The amount of heat gained by the water in cup 2 after adding the hot object(s) to it is 2119.121 Joules
Explanation:
As we know
Amount of heat gained
Q = mc (T2-T1)
Here,
mass of water in cup 2 (m) = 79.10 grams
Temperature of water in cup 2 = 16.8 degree Celsius
Specific heat of water (c) = 4.186 J/(g °C)
Final Temperature of water in cup 2 = 23.2 degree Celsius
Substituting the given values, we get -
Q = 79.10 * 4.186 * (23.2 -16.8) = 2119.121 Joules
The amount of heat gained by the water in cup 2 after adding the hot object(s) to it is 2119.121 Joules
They have a complete electronic configuration .their P shell is p6 therefore cannot gain or loose electron.
The acronym is RNA. Ribo Nucleic Acid
