22, 23, 24 , 25 please rate - k [NO-l IFl ch mechanism is consistent with the ra

Question

22, 23, 24 , 25 please

rate - k [NO-l IFl ch mechanism is consistent with the rate law? 2N02(g) + F2(g)# 2NO,F(g) 0.200 M NaOH to titrate to a phenolphthalein endpottll. Determine the molar concentration of the HCI solution. (A) 0.0830 M (C) 0.200 M (B) 0.105 M (D) 0.482 M NO2(g) + F(g) NO,F(g) F14) ? 2F(g) 2N018) + 2F(g) 2NO,F(g) NO2(g) + F2(g) NO,F(g) + F(g) 22. Calculate the [H,0'1 of a 0.10 M solution of hydrocyanic slow) (fast) (slow) (slow) acid, HCN. K,-4.9 x 10-1 (A) 0.10 M (C) 4.9 x 10-10 M (B) 7.0 x 106 M (D) 4.9 × 10-11M 23. What is the value of K., for this reaction at 25 °C? 2SOg)2SO2(8) +O2l) AG 141.7 kJ, 25 °C (A) 141.7 (C) 0.98 (fast) (B) 57 (D) 1.5×10-25 oose the potential energy diagram that has an tivation energy of 40 kJ and an overall energy change -100 kJ. 160 160 24. Which diagram represents represents an the most dilute solution of strong acid? Note: Water molecules are not shown. 120 undissociated acid (B) 120 Reaction Coordinate Reaction Coordinate 160 160 C 120 (D) 120 Reaction Coordinate Reaction Coordinate 25. For the reaction A ? B, which diagram o this information to answer 19 and 20. The figures nt the initial number of molecules in reaction vessels of depicts the equilibrium mixture of A and B if Keq-2.5? olumes and equal temperatures. A + B product rate KIAIIB edict the relative initial rates of reaction in vessels I, and III. ) II>I>III (B) III>I> II (C) I> III > II I, II, III would have the same rate. e relative values for the rate constant (k) for each case These all have the same rate constant

Explanation / Answer

22)

HCN dissociates as:

HCN -----> H+ + CN-

0.1 0 0

0.1-x x x

Ka = [H+][CN-]/[HCN]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((4.9*10^-10)*0.1) = 7*10^-6

since c is much greater than x, our assumption is correct

so, x = 7*10^-6 M

So, [H+] = x = 7*10^-6 M

Answer: B

23)

T= 25.0 oC

= (25.0+273) K

= 298 K

?Go = 141.7 KJ/mol

?Go = 141700 J/mol

use:

?Go = -R*T*ln Kc

141700 = - 8.314*298.0* ln(Kc)

ln Kc = -57.1931

Kc = 1.5*10^-25

Answer: D

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