Answer:
-68.4 kJ
Explanation:
<u>The standard enthalpy of vaporization = 23.3 kJ/mol</u>
<u>which means the energy required to vaporize 1 mole of ammonia at its boiling point (-33 °C).</u>
To calculate heat released when 50.0 g of ammonia is condensed at -33 °C.
This is the opposite of enthalpy of vaporization which means that same magnitude of heat is released.
<u>Thus, Q = -23.3 kJ/mol</u>
<u>Where negative sign signifies release of heat</u>
Given: mass of 50.0 g
Molar mass of ammonia = 17.034 g/mol
Moles of ammonia = 50.0 /17.034 moles = 2.9353 moles
Also,
1 mole of ammonia when condenses at -33 °C releases 23.3 kJ
2.9412 moles of ammonia when condenses at -33 °C releases 23.3×2.9353 kJ
<u>Thus, amount of heat released when 50 g of ammonia condensed at -33 °C= -68.4 kJ, where negative sign signifies release of heat.</u>
Arm. The center is the yellow, in the very middle. I hope this helps.
C) 3 Moles
This is because of the molecular structure shown in the image
Answer:
0.2
Explanation:
Given parameters:
Mass of helium = 0.628g
Mass of neon = 11.491g
Mass of argon = 7.613g
Unknown:
Mole fraction of neon = ?
Solution:
The mole fraction of an element is the number of moles of that element to the total number of moles in the gas mixture.
We need to calculate the number of moles of each element first;
Number of moles = 
Molar mass of Helium = 4g/mol
Molar mass of Neon = 20g/mol
Molar mass of Argon = 40g/mol
Number of moles of He =
= 0.16moles
Number of moles of Ne =
= 0.58moles
Number of moles of Ar =
= 0.19moles
Total number of moles = 0.16moles + 0.58moles + 0.19moles = 0.93moles
Mole fraction Neon =
= 0.2