Please show me the work, how you get it. Thank you 6/ consider the reaction 2h2
ID: 882106 • Letter: P
Question
Please show me the work, how you get it. Thank you
6/ consider the reaction 2h2 + O2 ---> 2h2O. What is the ratio of the initial rate of appearance of water to the initial rate of disappearance of oxygen?
A/ 1:1
B/ 2:1
C/ 1:2
D/ 3:2
7/ consider the following rate law: Rate = K[A]^n [B]^m. How are the exponents n and m determined ?
A/ By using the balanced equation
B/ By using the subcripts for the chemical formulas
C/ By hypothesis
D/ By experiment
8/ the catalyzed pathway in a reaction mechanism has a _______ activation energy and thus causes a ______ reaction rate
A/ higher, lower
B/ higher, higher
C/ lower, higher
D/ Lower, lower
16/ the rate law for the reaction A + 3B ----> C +2D is Rate =K[A]^2[B]^3. The order of the reaction is ?
A/ 2
B/ 3
C/ 5
D/ 6
17/ A Solution is made by dissolving 500 grams of ammonium nitrate in enough water to make 2.50 liters of solution. Determine the molarity and write the appropriate units.
26/ which of the following is a conjugate acid/base pair?
A/ HCL/OCL-
B/ H2SO4/SO4^2-
C/ NH4+/ NH3
D/ H3O+/OH-
28/ A proton in an aqueous solution combines with a water molecule to form a
A/ H+
B/H3O+
C/ OH-
D/ OH2-
29/ which pair acts as bronsted – lowry bases in the equilibrium:
HCL + NH3 ---> <---- NH4+ CL-
A/ HCL and NH4+
B/ NH3 and CL-
C/ HCL and CL-
D/ NH3 and NH4+
31/ 2NO + H2 ----> N2O + H2O
experiment initial[NO] (mol/L) initial [H2] ( mol/L) rate ( mol/L*S)
1 6.4 x 10^-3 2.2 x 10^-3 2.6 x 10^-5
2 12.8 x 10^-3 2.2 x 10^-3 1.0 x 10^-4
3 6.4 x 10^-3 4.5 x 10^-3 5.1 x 10^-5
I/ write the rate law for this reaction
II/ Determine the rate constant for this reaction with the appropriate units/
III/ what is the order of this reaction?
36/ Write the equilibrium expression Kc for the reversible reaction:
2NO(g) + O2(g) -----> 2NO2(g)
37/Initially 2.0 moles of N2(g) and 4.0 moles of H2(g) were added to a 1.0-liter container and the following reaction then occurred:
3H2(g) + N2(g) 2NH3(g)
The equilibrium concentration of NH3(g) = 0.68 moles/liter at 700°C. Determine Kc at 700°C for the formation of ammonia
38/ For the reaction: H2(g) + Br2(g) ----> 2HBr(g) kp= 3.5 x 10^4 at 1495k
what is the value of Kp for the following reaction at 1495K.
I/ 1/2H2(g) + 1/2Br2(g) ----> HBr(g)
II/ 4HBr(g) -----> 2H2(g) + 2Br2(g)
41/ calculate the [H+] in a solution that has a pH of 11.70 at 25 celcius.
42/ Determine the pH of a solution at 25 celcius in which the [OH-] = 3.4 x 10^-5 M.
43/ Calculate the pH of a 0.250 M HNO3(aq) solution at 25 celcius.
44/ Determine the pH of a solution at 25 celcius if the pOH is 3.75.
45/ Determine the molarity of a solution prepared by adding 350 grams of potassium hydroxide to enough water to make 5.75 liters of solution.
47/ of the following, a 0.10M aqueous solution of __________ will have the highest boiling point.
A/ NACL
B/ AL(NO3)3
C/ K2CrO4
D/ Na2SO4
48/ Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction:
2NO2 -----> 2NO + O2
In a particular experiment at 300 celcius , [NO2] drops form 0.0100 M to 0.00650 M in 100 seconds. The rate of appearance of O2 for this period is _________ M/s.
A/ 1.8 x 10^-5
B/ 3.5 x 10^-5
C/ 7 x 10^-5
D/ 3.5 x 10^-3
E/ 7 x 10^-3
Explanation / Answer
6)1/2 RATE OF DISAPPEARENCE OF NO2 = RATE OF APPEARENCE OF O2
(RATE OF DISAPPEARENCE OF NO2)/(RATE OF APPEARENCE OF O2) = 2/1
ANSWER IS B 2:1
7) D) BY EXPERIMENT
8) C) lower, higher
16) C) 5
SUM OF THE POWERS OF CONCENTRATION TERMS INVOLCED IN THE RATE LAW EXPRESSION IS CALLED AS ORDER OF THE REACTION.
17)
M= (wt.of ammonium nitrate)/(m,wt of ammonium nitrate*volume)
= 500/(80.050*2.50)
= 2.49M
26) C) NH4^+ & NH3 BECAUSE ONLY ONE PROTON DIFFERENCE
28) B) H3O^+
H^+ + H2O ----------------------------> H3O^+
29) B) NH3 AND Cl^-1 because both are accepting proton
36) Kc = [NO2]2/[NO]2[O2]1
45) B.P OF A SOLUTION IS DIRECTLY PROPOTIONAL TO THE i*m
answer is B Al(NO3)3 BECAUSE i=4 , Al+3 + 3 NO3-
48) 1/2 RATE OF DISAPPEARENCE OF NO2 = RATE OF APPEARENCE OF O2
RATE OF APPEARENCE OF O2 = (1/2)*((0.0100-0.00650)/100)
RATE OF APPEARENCE OF O2 = 1.75*10^-5
45)
M= (wt.of KOH)/(m,wt of KOH*volume)
= 350/(56.10*5.75)
= 1.085M
43) P^H = -log[H^+]
HNO3---------------------------> H^+ NO3^-1
0.250M 0.250 0.250
P^H = -log(0.250)
= 0.602
44) P^H + P^OH = 14
P^H = 14- P^OH = 14-3.75 = 10.25
42) P^OH = -log[OH^-1]
= -log (3.4*10^-5)
= 4.46
p^H = 14-P^OH = 14-4.46 = 9.53
41)
P^H = -log[H^+]
11.70 = -log[H^+]
[H^+ ]= 1.99*10^-12