{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Equation Sheet

# Equation Sheet - Useful equations AE=q W PV=nRT...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Useful equations AE=q+W PV=nRT 131=[A]‘ _kt w=-PAV an2 1‘11“ 10- (P+—2)(V—nb)=nRT AH=AE+PAV V 1=kr+ 1 AG°=AH°-TAS° PT=P1+P2+P3... [A] . [A10 AG° = -RT In K P1 = mm [A]: = _kt + [A10 AG=AQO+RTan v: 3RT 1 =sphtx-mxAT M t1/2=k[A]0 3.. _ An ratel _ £2 KP—KCRT) mt“2 _ “‘1 0.693 KCZkf/kr 2=c/v I t”— k E OH=-lo OH' A pH= l gﬂgﬂl E=(hxc)/A k: Ae RT P ' 0g E=1/2mv2 ﬂ: EA(_I___1_) PH+pOH=pKw E=mc2 k2 R T 75 pKw=-long O l=h/mv Pl AH 1 1 pKa = Jog Ka 1n}— — R (F _ F) PKb_ logK (AX)(Amv)2h/ 4“ 2 1 2 "' b _. 1 l 1 ATf:KmeXl K”XK"’K“ TR" 73772 AT —K [H+][0H‘] =Kw= 1.0x 10'14 b— 13X m X: H=MRTi Avogadro’s number ‘ Gas constant Charge on an electron Planck constant Boltzmann constant Rydberg constant Fundamental Constants Mass of an electron at rest Speed of light in vacuum Acceleration due to gravity NA = 6.02214 x 1023 mor‘ .R = 0.082057 liter atm K“ mor‘ = 8.3145 JK" 11101" = —1.60218 x10'l9 c me = 9.1094 x 10‘31 kg Conversion Factors 1 oz = 28.35 g 1 kg = 2.205 lb 1 amu = 1.66054 x 10‘27 kg 1 atm ‘ = 760 torr = 101,325 Pa = 6.62607 x 10—34 J s = 6.62607 x 10'34 kg m2 s'1 = 2.99792 x 108 m s" = 9.80665 m 3'2 = 1.38065 x 10‘23 JK" R = 1.0973732 x 10'2 nm1 = 4.184 J leV = 1.60218x10‘191 1 eV = 96.485 kJ mol‘1 lL.atm = 101.325 J 0°C =273.15 K ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern