The answer is (4) at the cathode, where reduction occurs. The Na+ gains one electron and become Na(l). So the reaction occurs at cathode and is reduction reaction.
Solution :
From the balanced chemical equation, we can say that 1 moles of KBr will produce 1 moles of KCl .
Moles of KBr in 102 g of potassium bromide.
n = 102/119.002
n = 0.86 mole.
So, number of miles of KCl produced are also 0.86 mole.
Mass of KCl produced :
Hence, this is the required solution.
During a phase change the temperature does not change since all of the heat is being absorbed in order to break the intermolecular forces. Due to that, the formula will not need to have T in it and is actually q=nΔH(v).
n=the number of moles (in this case 2.778mol of water since you divide 50g by 18g/mol).
ΔH(v)=the molar heat of vaporization (in this case 40.7kJ/mol).
q=the heat that must be absorbed
q=2.778mol×40.7kJ/mol
q=113.1kJ
Therefore the water needs to absorb 1.13×10²kJ.
I hope this helps. Let me know if anything is unclear.
Answer:
B) irreversible process
Explanation:
The process given here is irreversible.
DeltaH formation = deltaH of broken bonds - deltaH of formed bonds
Broken bonds: tiple bond N-N and H-H bond
Formed bonds: N-H and N-N bonds
You also have to take note of the molar coefficients
deltaH formation = <span> [(N≡N) + 2 * (H-H)] - [4 * (N-H) + (N-N)]
= (945 + 2*436) - (4*390 + 240)
= 17 kJ/mol
The answer is 17 kJ/mol.</span>
