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>
Answer:
AgNO3 -soluble in water
AgCl- insoluble in water
AgI- Insoluble in water
Explanation:
The solubility of chemical compounds in water is easily predicted by a set of rules generally referred to as the solubility rules. These rules are usually based on experimental observation of diverse groups of compounds.
According to the solubility rules, nitrates are soluble in water including the nitrates of silver. However, halides are soluble in water except those of silver, lead and mercury.
Single replacement reactions. For example copper is more reactive than silver. So a copper wire in a silver solution will cause the silver to become a metal again.
Answer:
It is either the third or fourth statement.
Explanation:
This is because exothermic reactions give off heat.
