Recent content by Hemolymph

oh i did it wrong then i just assumed it was .1 because 10%=.1 same for 90% being .9

(1/.9)=1/.1+k(20 seconds) I got k to be .005555 so If I find t i get (1/.1)=(1/.9)+.005555(t) 9/.0055555=t t=1600s that look like the right path to get to the answer?

Would I just be able to just substitute the percentages in as if they were concentrations?

Do you know how i could tackle the second one ?

ok so I did ln(Rate 2/rate1)=71/.008314-((1/423)-(1/443)) got .911452 took e^.911452 = 2.5 so 2.5 times greater is rate 2 to rate 1

ln(k1/k2)=Ea/R(1/t_2)-(1/t_1)?

I can't see how I can apply the Arrhenius equation if I don't have any activation energies

Ah ok thanks

Homework Statement I have two questions that I just don't even know where to start The activation energy for the reaction CH3CO CH3 + CO is 71 kJ/mol. How many times greater is the rate constant for this reaction at 170°C than at 150°C? A) 0.40 B) 1.1 C) 2.5 D) 4.0 E) 5.0 and A...

Ok, so I did that I got the Concentration of of copper cyanide to be 1.28E7 then did the reverse rxn CuCN_3_→3CN+Cu (1.28E7) 3x x the new K value i got to be 1/1E9=1E-9 so (27x^4/(1.28E7))=1E-9 i got x to be .0147 The answer booklet says 1.3E-14

Homework Statement 50.0 mL sample of 2.0 10–4 M CuNO3 is added to 50.0 mL of 4.0 M NaCN. The formation constant of the complex ion Cu(CN)3 is 1.0 10^9. What is the copper(I) ion concentration in this system at equilibrium? The Attempt at a Solution I did a dilution and got the new...

Homework Statement . You have solutions of 0.200 M HNO2 and 0.200 M KNO2 (Ka for HNO2 = 4.00 × 10–4). A buffer of pH 3.000 is needed. What volumes of HNO2 and KNO2 are required to make 1 liter of buffered solution? A) 500 mL of each B) 286 mL HNO2; 714 mL KNO2 C) 413 mL HNO2; 587 mL KNO2...

Thanks for the advice/ help

I was doing a simliar problem of [-5,0] where it was evaluating 1+rad(25-x^2) dx And the solution had a the area broken up into a rectangle and a semicircle I guess i tried to apply the same technique to this problem. The answer came out to be 5+(25pi)/4

Homework Statement from [-2,2] rad(4-x^2) Homework Equations The Attempt at a Solution I know its a circle and i get the equation to be y^2+x^2=4 and I believe it has to be divided into a circle and rectangle so the area of the rectangle i got to be 2 the circle i got to be...