To calculate the <span>δ h, we must balance first the reaction:
NO + 0.5O2 -----> NO2
Then we write all the reactions,
2O3 -----> 3O2 </span><span>δ h = -426 kj eq. (1)
O2 -----> 2O </span><span>δ h = 490 kj eq. (2)
NO + O3 -----> NO2 + O2 </span><span>δ h = -200 kj eq. (3)
We divide eq. (1) by 2, we get
</span>O3 -----> 1.5O2 δ h = -213 kj eq. (4)
Then, we subtract eq. (3) by eq. (4)
NO + O3 -----> NO2 + O2 δ h = -200 kj
- (O3 -----> 1.5 O2 δ h = -213 kj)
NO -----> NO2 - 0.5O2 δ h = 13 kj eq. (5)
eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)
O -----> 0.5O2 <span>δ h = -245 kj eq. (6)
</span>
Add eq. (6) to eq. (5), we get
NO -----> NO2 - 0.5O2 δ h = 13 kj
+ O -----> 0.5O2 δ h = -245 kj
NO + O ----> NO2 δ h = -232 kj
<em>ANSWER:</em> <em>NO + O ----> NO2 δ h = -232 kj</em>
The answer is (2). If you recall Rutherford's gold foil experiment, remember that a stream of positively charged alpha particles were shot at a gold foil in the center of a detector ring. The important observation was that although most of the particles passed straight through the foil without being deflected, a tiny fraction of the alpha particles were deflected off the axis of the shot, and some were even deflected almost back to the point from which they were shot. The fact that some of the alpha particles were deflected indicated a positive charge (because same charges repel), and the fact that only a small fraction of the particles were deflected indicated that the positive charge was concentrated in a small area, probably residing at the center of the atom.
The answer is number 2. That releases massive amounts of radiation and by the way, that is how atomic bombs are made to detonate.
Answer:
C. A hydrocarbon molecule containing six carbon atoms and only
single bonds
Explanation:
hope it helps
I suppose that the answer is A
