The volume of gas at temperature of 20 ∘ C and pressure of 760 mmHg has to be calculated Concept Introduction: Charles’s Law: Boyle’s law states that “ at a constant pressure (P), the volume (V) occupied by any sample of gas is directly proportional to the absolute temperature (T)’’ V α T (at constant P) For a sample of gas supposing a change in temperature, Charles’s law can be written as V f T f = V i T i (at constant P) On rearranging, we get V f = V i × T f T i (at constant P) Where, T i a n d T f are initial temperature and final temperature respectively V i a n d V f are initial volume and final volume respectively
The volume of gas at temperature of 20 ∘ C and pressure of 760 mmHg has to be calculated Concept Introduction: Charles’s Law: Boyle’s law states that “ at a constant pressure (P), the volume (V) occupied by any sample of gas is directly proportional to the absolute temperature (T)’’ V α T (at constant P) For a sample of gas supposing a change in temperature, Charles’s law can be written as V f T f = V i T i (at constant P) On rearranging, we get V f = V i × T f T i (at constant P) Where, T i a n d T f are initial temperature and final temperature respectively V i a n d V f are initial volume and final volume respectively
Solution Summary: The author explains how the volume of gas at a constant pressure (P) is directly proportional to the absolute temperature (T).
The volume of gas at temperature of 20∘C and pressure of 760 mmHg has to be calculated
Concept Introduction:
Charles’s Law:
Boyle’s law states that “at a constant pressure (P), the volume (V) occupied by any sample of gas is directly proportional to the absolute temperature (T)’’
Vα T (at constant P)
For a sample of gas supposing a change in temperature, Charles’s law can be written as
VfTf=ViTi (at constant P)
On rearranging, we get
Vf=Vi×TfTi (at constant P)
Where,
Tiand Tf are initial temperature and final temperature respectively
Viand Vf are initial volume and final volume respectively
For a titration of 40.00 mL of 0.0500 M oxalic acid H2C2O4 with 0.1000 M KOH, calculate the pH at each of the following volume of KOH used in the titration: 1) before the titration begin;2) 15 mL; 3) 20 mL; 4) 25 mL; 5) 40 mL; 6) 50 mL. Ka1 = 5.90×10^-2, Ka2 = 6.50×10^-5 for oxalic acid.
Predict the major organic product(s), if any, of the following reactions. Assume all reagents are in excess unless otherwise indicated.
Chapter 5 Solutions
OWLv2 for Ebbing/Gammon's General Chemistry, 11th Edition, [Instant Access], 1 term (6 months)
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell