If you were to engineer an everyday solution for conserving water which activity do you think would be the most impactful ?
Answer:
Explanation:
= Initial pressure = 931 torr = 
= Final pressure = 113 kPa
= Initial volume = 350 mL
= Final volume
From the Boyle's law we have
The volume the gas would occupy is .
Answer:
P = 27.9 atm
Explanation:
Given data:
Mass of CO₂ = 25 g
Temperature = 25°C (25+273.15 K = 298.15 K)
Volume of gas = 0.50 L
Pressure of gas = ?
Solution:
Firs of all we will calculate the number of moles of gas,
Number of moles = mass/molar mass
Number of moles = 25 g/ 44 g/mol
Number of moles = 0.57 mol
Pressure of gas :
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
P × 0.50 L = 0.57 mol × 0.0821 atm.L/ mol.K × 298.15 K
P = 13.95 atm.L/ 0.50 L
P = 27.9 atm
Answer:
Boiling point for the solution is 100.237°C
Explanation:
We must apply colligative property of boiling point elevation
T° boiling solution - T° boiling pure solvent = Kb . m
m = molalilty (a given data)
Kb = Ebulloscopic constant (a given data)
We know that water boils at 100°C so let's replace the information in the formula.
T° boiling solution - 100°C = 0.512 °C/m . 0.464 m
T° boiliing solution = 0.512 °C/m . 0.464 m + 100°C → 100.237 °C
