Search found 5 matches
- Fri Mar 11, 2016 9:39 pm
- Forum: Phase Changes & Related Calculations
- Topic: Winter 2011 Final 1A: Heat required to boil water
- Replies: 3
- Views: 743
Re: Winter 2011 Final 1A: Heat required to boil water
Step 1 : Heat liquid water from 25.0°C to 100°C (q 1 ) q=mC s ΔT q 1 =(50.0 g)(4.184 J.°C -1 .g -1 )(75.0 °C) q 1 =15.69 kJ Step 2 : Turn liquid water to water vapor (q 2 ) ΔH vap =40.7 kJ.mol -1 (from equation sheet) 50.0 mL=50.0 g=2.775 mol of water (convert given mL to mol) q=mΔH vap q 2 =(2.775...
- Mon Mar 07, 2016 10:11 pm
- Forum: Phase Changes & Related Calculations
- Topic: Winter 2011 Final 1A: Heat required to boil water
- Replies: 3
- Views: 743
Winter 2011 Final 1A: Heat required to boil water
Q: Suppose you want to use methane gas (ΔH c =-890. kJ/mol) to boil 50.0 mL of water at 25.0°C. How many grams of methane gas would need to be combusted to generate the required heat? A: Solve for q (heat) using q=mC s ΔT, which gives q=15.7 kJ. Then, use q=nΔH c to find the moles of methane, and fi...
- Wed Jan 13, 2016 5:21 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 3874333
Re: Chemistry Jokes
polar bears...
- Fri Jan 08, 2016 6:59 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: 8.10 units question
- Replies: 3
- Views: 782
Re: 8.10 units question
If you want, you can use 1 J=.009869 L*atm as a shortcut to convert L*atm to joules! The value is derived from the gas constant, R. (Divide R=8.314 J/(K*mol) by R=8.206x10^-2 L*atm*K^-1, then divide 1 by that quotient to get 1 J=.009869 L*atm)
- Wed Dec 02, 2015 9:20 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Video: Quiz 3 Problem 8
- Replies: 3
- Views: 747
Re: Video: Quiz 3 Problem 8
A question about chemical equilibrium from self-prep quiz 3