Answer:
The heat of vaporisation of methanol is "3.48 KJ/Mol"
Explanation:
The amount of heat energy required to convert or transform 1 gram of liquid to vapour is called heat of vaporisation
When 8.7 KJ of heat energy is required to vaporize 2.5 mol of liquid methanol.
Hence, for 1 mol of liquid methanol, amount of heat energy required to evaporate the methanol is = 
= 3.48 KJ
So, the heat of vaporization 
Therefore, the heat of vaporization of methanol is 3.48KJ/Mol
The grams of 22.9 % sugar solution that contain 68.5 g of sugar is 299.13 g of solution
<u><em>calculation</em></u>
22.9% means that there are 22.9 g of sugar in 100 g of solution.
what about 68.5 g of sugar
- <em>by cross multiplication</em>
=[(68.5 g sugar x 100 g solution) /22.9 g sugar] =299.13 g of solution
Nb; <em>g sugar cancel each other</em>
Answer:
Start and end times; distance run.
Step-by-step explanation:
Average speed = distance/time.
Kaila should record the distance run, the time she started, and the time she ended her run.
The difference between the start and end times gives the time for the run.
If she inserts her numbers into the formula, she will get her average running speed.
212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
Option A.
Explanation:
Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.
So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.
Then, 
So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
Mass of NH₃ produced = 34 g
Explanation:
Given data:
Mass of nitrogen = 28 g
Mass of Hydrogen = 12 g
Mass of NH₃ produced = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Moles of nitrogen:
Number of moles = mass/molar mass
Number of moles = 28 g/ 28 g/mol
Number of moles = 1 mol
Moles of hydrogen:
Number of moles = mass/molar mass
Number of moles = 12 g/ 2 g/mol
Number of moles = 6 mol
Now we will compare the moles of hydrogen and nitrogen with ammonia.
H₂ : NH₃
3 : 2
6 : 2/3×6 = 4 mol
N₂ : NH₃
1 : 2
Number of moles of ammonia produced by nitrogen are less thus it will act as limiting reactant.
Mass of ammonia produced:
Mass = number of moles × molar mass
Mass = 2 mol × 17 g/mol
Mass = 34 g
