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q = + 6.5 * 10^-6 C ; Electric field points away from the charge.  E = k q / r^2 = 8.99*10^9 * 6.5 * 10^-6 / 0,39^2 = 3,8 * 10^5 N/ C . It points outward because the charge is positive. Ciao

Mark. Marc feels a bit pretentious to me for some reason? Like I imagine a flashy guy who talks a lot about his expensive car. 

Air miles: About 5,925 km. (Agua Prieta to Lima) One checkpoint (customs and immigration in Peru) Not hard if you have a passport. Driving miles: Something in excess of 6,250 km. You must pass...

The integrated rate law for a 2nd order reaction is: 1/[B] = kt + 1/[B]o 1/[B] = 2.45X10^-2 M-1min-1 (1.8 min) + 1/(6.9 M) [B] = 5.3 M

Won't use there notation  n_ C _ k    , I will write as    C (n, k )       so both...7!) = 10*9*8/ 3*2*1  =   10*9*4/3  = 10*3*4  = 120  C (10,3) (1/4)^3 (3/4)^7 =  120* ( 3^7/ 4^10)  =  = 262440...

(8.49∘ C ) / (37.8 ∘ C / m) = 0.22460 m (0.22460 mol/kg) x (0.0250 kg) = 0.005615 mol (2.53 g) / (0.005615 mol) = 450.5788 g/mol = 451 g/mol

K = √s(s-a)(s-b)(s- c ) where s = (1/2) (a + b + c ) s= (1/2)(100 + 50 +75) = 112.5 K = √s(s-a)(s-b)(s- c )  K = √112.5(112.5 -100)(112.5 -50)(112.5 - 75) = 1815.46 1815.46 x $ 3.00 = $. 5446.38

...) / ((0.08206 L⋅atm/(mol⋅ K )) × (29 + 273) K ) = 0.0119219 mol = 0.01192 mol H2 So you are correct about Parts a), B), and C ). Part D) Given: x + y = 0.250...