A displacement reaction will occur from the system given above. The chlorine molecules will displace the bromide ions in the solution of sodium bromide. The reaction will yield to sodium chloride and bromine. The reaction will be:
2NaBr + Cl2 = 2NaCl + Br2
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>
P1v1=p2v2
47.5*125=66.2*v2
v2= p1v1/p2
=(47.5)(125)/(66.2)
=89.69ml
To solve this, we should follow order of operations. To start, we should multiply the values inside of the parentheses.
(34.6785*5.39)+435.12
186.917115+435.12
Now, we should add the 2 values we are left with together.
186.917115
<span><u>+435.120000</u>
</span> 622.037115
Using the math above, we can see that this expression is equal to 622.037115.
The answer to the question is false
