C6H6 + 02 forms CO2 + H20
Complete with factors to stabilize
C6H6 + 15/2 O2 forms 6 CO2 + 3 H2O
You take away only the info you need
C6H6 15/2 O2
1 mol 15/2 or 7,5 mol
15/2 or 7,5 mol of O2 are required.
;)
<span>Answer:
A 1.00 L solution containing 3.00x10^-4 M Cu(NO3)2 and 2.40x10^-3 M ethylenediamine (en).
contains
0.000300 moles of Cu(NO3)2 and 0.00240 moles of ethylenediamine
by the formula Cu(en)2^2+
0.000300 moles of Cu(NO3)2 reacts with twice as many moles of en = 0.000600 mol of en
so, 0.00240 moles of ethylenediamine - 0.000600 mol of en reacted = 0.00180 mol en remains
by the formula Cu(en)2^2+
0.000300 moles of Cu(NO3)2 reacts to form an equal 0.000300 moles of Cu(en)2^2+
Kf for Cu(en)2^2+ is 1x10^20.
so
1 Cu+2 & 2 en --> Cu(en)2^2+
Kf = [Cu(en)2^2+] / [Cu+2] [en]^2
1x10^20. = [0.000300] / [Cu+2] [0.00180 ]^2
[Cu+2] = [0.000300] / (1x10^20) (3.24 e-6)
Cu+2 = 9.26 e-19 Molar
since your Kf has only 1 sig fig, you might be expected to round that off to 9 X 10^-19 Molar Cu+2</span>
The concentration of the drug stock solution is 1.5*10^-9 M i.e. 1.5 * 10^-9 moles of the drug per Liter of the solution
Therefore, the number of moles present in 1 ml i.e. 1*10^-3 L of the solution would be = 1 *10^-3 L * 1.5 * 10^-9 moles/1 L = 1.5 * 10^-12 moles
1 mole of the drug will contain 6.023*10^23 drug molecules
Therefore, 1.5*10^-12 moles of the drug will correspond to :
1.5 * 10^-12 moles * 6.023*10^23 molecules/1 mole = 9.035 * 10^11 molecules
The number of cancer cells = 2.0 * 10^5
Hence the ratio = drug molecules/cancer cells
= 9.035 *10^11/2.0 *10^5
= 4.5 * 10^6
273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit
