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>
A chemical reaction must occur for a compound to have different properties.
Answer: Hello the compound is missing but I was able to get the Full question and missing compound . ( compound = copper sulfate )
<em>answer</em> : statement ; 2 , 3 and 5
Explanation:
The true statements regarding the coordination compound ( copper sulfate ) are :
- The ligand must have at least one unshared pair of valence electrons in order to covalently bond with transition metal in the coordination compound ( statement 2 )
- Ethanol was used during crystallization of the coordination compound because the compound is soluble in ethanol ( statement 3 )
- The colors of many coordination compounds are the result of light absorption by the d electrons on the transition metal ( statement 5 )
During the coordination of compounds dative bonds exits between the transition metals and the Ligands molecules
Don’t really understand what you’re asking but, if you’re asking how to read a graduated cylinder:
Look at the graduated cylinder at eye level, find the meniscus, whatever the meniscus is at is your answer.
