Answer:
121 K
Explanation:
Step 1: Given data
- Initial volume (V₁): 79.5 mL
- Initial temperature (T₁): -1.4°C
- Final volume (V₂): 35.3 mL
Step 2: Convert "-1.4°C" to Kelvin
We will use the following expression.
K = °C + 273.15 = -1.4°C + 273.15 = 271.8 K
Step 3: Calculate the final temperature of the gas (T₂)
Assuming ideal behavior and constant pressure, we can calculate the final temperature of the gas using Charles' law.
V₁/T₁ = V₂/T₂
T₂ = V₂ × T₁/V₁
T₂ = 35.3 mL × 271.8 K/79.5 mL = 121 K
Answer:

Explanation:
Given:
For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.
To find: fraction of the whole athletic field reserved for each fifth class
Solution:
Fraction of the whole athletic field reserved for four fifth classes = 
So, fraction of the whole athletic field reserved for each fifth class = 
Chloroacetic acid is stronger than acetic acid because of the electron-withdrawing effect of chlorine. This effect is caused by the electronegativity.
Answer:
Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em> by adding 1 mol of
to 1 kg of water.
Explanation:
1) Moles of NaCl ,
Mass of water = m= 1 kg = 1000 g
Moles of water = 
Vapor pressure of the solution = 
Vapor pressure of the pure solvent that is water = 
Mole fraction of solute(NaCl)= 



The vapor pressure for the NaCl solution at 17.19 Torr.
2) Moles of sucrose ,
Mass of water = m = 1 kg = 1000 g
Moles of water = 
Vapor pressure of the solution = 
Vapor pressure of the pure solvent that is water = 
Mole fraction of solute ( glucose)= 



The vapor pressure for the glucose solution at 17.19 Torr.
p = p' = 17.19 Torr
Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of
to 1 kg of water.