A) Particles of gas move slower.
B) Gas changes to liquid.
C) The gas loses thermal energy.
D) Gas particles decrease.
Answer: 1.56 ATM
Explanation: if we assume temperature is constant, gas obeys
Boyles law pV= constant. Then p1·V1= p2·V2. And V1 = p2V2/p1
= 3.0 atm·0,52 l / 1.0 atm
I believe the answer you're looking for is B. It's Polluted. I hope this helps.
Answer:
∆H° rxn = - 93 kJ
Explanation:
Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.
We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.
N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)
1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)
∆H° rxn = ∑ H bonds broken - ∑ H bonds formed
∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]
∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ
be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .
The trough and the hill part
