Sort by

- Relevance
- |Time

...temperature is decreased , then K will increase from its original

**K**(**c**) [ when temperature is normal ] so By Le-chatelier's principle ,**K**...1 Answers · Science & Mathematics · 09/07/2016

V1 / T1 = V2 / T2 Substitute and solve. Don't forgot to convert deg

**C**to**K**for gas problems.2 Answers · Science & Mathematics · 13/07/2016

...0 K = -273.15°C and 273.15 K = 0°C; that is,

**K**= °**C**+ 273.15 You can think of both temperature scales having the same...3 Answers · Science & Mathematics · 02/07/2016

... the concentrations in equilibrium state satisfy the relation:

**K**= [**C**]/ ( [A]∙[B] ) Equilibrium constant and free energy change ...3 Answers · Science & Mathematics · 23/06/2016

The range when converted from degree celsius to Kelvin is 263 to 278

**K**. Add up 273 and (-10) and 273 with 5. U have probably understood the formula if not please comment me.Thank you1 Answers · Science & Mathematics · 16/05/2016

0.79 J/(g*

**K**) = 790 J/(kg***C**)3 Answers · Science & Mathematics · 25/04/2016

...251 L) / ((3.33 x 10^-3 mol) x (62.36367 L torr/K mol)) = 886

**K**(**c**) P = nRT / V = (0.0466 mol) x (0.08205746 L atm/**K**...1 Answers · Science & Mathematics · 15/04/2016

... / dS .. T = (92380 J) / (198.3 J/K) = ___

**K**= __ °**C********* you get to finish1 Answers · Science & Mathematics · 10/04/2016

After the reaction, C2H5OH and CH3COOH remained 1/3 mol of each, that meant the amounts of alcohol and acid used in this reaction are 2/3 mol of each. The equilibrium constant: kc = [CH3COOC2H5][ H2,O...

1 Answers · Science & Mathematics · 28/03/2016

**K**= °**C**+ 273.15. Hence, °C = K - 273.15 = 4 - 273.15 = -269.15° C2 Answers · Science & Mathematics · 17/03/2016