3 facts about transfer of enegry
2 answers:
You might be interested in
Answer:
The answer to your question is P = 1.357 atm
Explanation:
Data
Volume = 22.4 L
1 mol
temperature = 100°C
a = 0.211 L² atm
b = 0.0171 L/mol
R = 0.082 atmL/mol°K
Convert temperature to °K
Temperature = 100 + 273
= 373°K
Formula
Substitution
Simplify
(P + 0.0094)(22.3829) = 30.586
Solve for P
P + 0.0094 =
P + 0.0094 = 1.366
P = 1.336 - 0.0094
P = 1.357 atm
42.4 ml is the volume in milliliters of the lead ball if a lead ball is added to a graduated cylinder containing 50.6 ml of water.
<h3>What is a graduated cylinder?</h3>A tall narrow container with a volume scale is used especially for measuring liquids.
The graduated cylinder contains water
mL is a volume unit.
Water volume = 50.6 ml
The lead ball caused an increase in volume from 50.6 ml to 93.0 mL.
The new volume is the lead ball volume plus the original water volume :
Final volume = Vlead ball+ Water original volume
Hence, 42.4 ml is the volume in milliliters of the lead ball.
Learn more about the graduated cylinder here:
#SPJ1