Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: PCl5(g) → PCl3(g) + Cl2(g)

P4(s) + 6Cl2(g) → 4PCl3(g) ΔH = -2439 kJ
4PCl5(g) → P4(s) + 10Cl 2(g) ΔH = 3438 kJ

The value of for the desired reaction will be 249.75 KJ.

Explanation:

The desired reaction is shown below

The desired reaction can be obtained by adding the given reactions and then dividing both sides by 4.

Net Enthalpy change for the desired reaction

your answer would be “dry”

1. a)c-14 has two more protons

2. d)they have different numbers of protons, neutrons, and valence electrons; they are all gases at room temperature.

3. c) iron and nickel.