Answer:
V₂ = 16.5 L
Explanation:
To solve this problem we use <em>Avogadro's law, </em>which applies when temperature and pressure remain constant:
V₁/n₁ = V₂/n₂
In this case, V₁ is 22.0 L, n₁ is [mol CO + mol NO], V₂ is our unknown, and n₂ is [mol CO₂ + mol N₂].
- n₁ = mol CO + mol NO = 0.1900 + 0.1900 = 0.3800 mol
<em>We use the reaction to calculate n₂</em>:
2CO(g) + 2NO(g) → 2CO₂(g) + N₂(g)
0.1900 mol CO * 
0.1900 mol CO₂
0.1900 mol NO * 
0.095 mol N₂
- n₂ = mol CO₂ + mol N₂ = 0.1900 + 0.095 = 0.2850 mol
Calculating V₂:
22.0 L / 0.3800 mol = V₂ / 0.2850 mol
V₂ = 16.5 L
CH3NH2 can only have as many hydrogen bonds as hydrogen bonding sites in the molecule. CH3NH2 has two N−H bonds and a lone pair of electrons on the nitrogen atom. Therefore, CH3NH2 can form three hydrogen bonds with water.
Evaporation is a phase transition from the liquid phase to the gas phase that occurs at temperatures below the boiling point at a given pressure.
hope this helps :)
To find out how many grams are in 4.65 moles of Al(NO₂)₃
Find out what the molar mass of Al(NO₂)₃ is
Al = 26.98 g/mol Al
N = 14 g/mol N
O = 16 g/mol O
Next, you have to look at the subscripts and figure out which they belong to, in this case:
Al = 26.98 g/mol Al
N₃ = 42 g/mol N₃
O₆ = 96 g/mol O₆
Finally, add the numbers together, so:
26.98 g/mol Al + 42 g/mol N₃ + 96 g/mol O₆ =
164.98 g/mol Al(NO₂)₃
Now, you have 4.65 mol Al(NO₂)₃ so
164.98 g/mol Al(NO₂)₃ × 4.65 mol Al(NO₂)₃ =
767.157 grams of Al(NO₂)₃
Hydrogen bond is found between water molecule if im correct
