Answer:
1 may is the answer for the question
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
<h3>
Answer:</h3>
1.85 M
<h3>
Explanation:</h3>
<u>We are given;</u>
- Number of moles as 0.50 mol
- Volume of the solution is 270 ml
But, 1000 mL = 1 L
- Thus, volume of the solution is 0.27 L
We are required to calculate the molarity of the solution;
- Molarity refers to the concentration of a solution in moles per liter.
- It is calculated by dividing number of moles with the volume.
Molarity = Moles ÷ Volume
In this case;
Molarity = 0.50 moles ÷ 0.27 L
= 1.85 Mol/L or 1.85 M
Therefore, molarity of the solution is 1.85 M
Answer:
219.95 °C
Explanation:
Given data:
Volume of gas = 9.71 L
Initial pressure = 209 torr (209/760 = 0.275 atm)
Initial temperature = 10.1 °C (10.1 +273 = 283.1 K)
Final temperature = ?
Final pressure = 364 torr (364/760 =0.479 atm)
Solution:
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
0.275 atm / 283.1 K = 0.479 atm/T₂
T₂ = 0.479 atm × 283.1 K/ 0.275 atm
T₂ = 135.6 atm. K /0.275 atm
T₂ = 493.1 K
Kelvin to °C:
493.1 K - 273.15 = 219.95 °C
Answer:
c
Explanation:
because nothing is changing, so what are you gonna track with the line graph
