Answer:
The Kinetic energy decreases
Explanation:
because the speed of the molecules slows down.
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>
Reactants bonds are broken and products bonds are formed
Answer:
A) oxidizing agent is SO2
B) NaClO is the oxidizing agent
Explanation:
A) This is a redox reaction in which oxidation and reduction occur simultaneously.
Thus, in 2H2S(g) + SO2(g) -> 2H2O(l) + 3S(s);
H2S is reduced as follows;
H2S → S + 2H+ + 2e−
We can see that SO2 has been reduced while H2S gets oxidized since it has changed state from - 2 to 0 . Thus sulphur dioxide is the oxidizing agent.
B) SO2(g) + H2O(l) + NaClO(aq) -> NaCl(aq) + H2SO4(aq)
In this, SO2 undergoes oxidation and NaClO is the oxidizing agent
