HIO_3 behaves as acid in water HIO_3 (aq) IO_3^- (aq) + H^+ (aq), with K_c = 0.1
ID: 994807 • Letter: H
Question
HIO_3 behaves as acid in water HIO_3 (aq) IO_3^- (aq) + H^+ (aq), with K_c = 0.17 at 25 degree C. What is the H^+ concentration in a solution that is initially 0.50 M HIO_3? 0.34 M 0.29 M 0.22 M 0.28 M The atmospheric pressure in the eye of a hurricane is found to be 26.6 in. of mercury, which is equivalent to 760 mm Hg 1.00 Times 10^5 Pa. 0.987 bar. 0.889 atm. 0.785 atm Given the following data for the reaction A rightarrow B, determine the activation energy, E_a, of the reaction to three significant figures. 6.89 J/mol 54.4 J/mol 57.3 J/mol 39.9 J/mol Oxidation refers to a gain in the number of protons an increase in oxidation number an increase in mass a decrease in oxidation number. What is the change in internal energy (Delta E) of a system when it is heated with 35.0 J of energy while it does 15.0 J or work? +35.0 J -20.0 J +50.0 J +20.0 J The half-life of radioactive carbon-14 is 5730 years. If the^14 C level in a sample of organic matter has been reduced to 0.200% of its original value, approximately how much time has passed? Radioactive decay follows first-order kinetics. 51,400 years 29,900 years 1650 years 9220 years Processes are always spontaneous when (H and S refer to the system). Delta H > 0 and Delta S > 0 Delta HExplanation / Answer
Solution :-
Q14
HIO3 ------- > H+ + IO3^-
0.50 0 0
-x +x +x
0.50-x x x
Kc= [H^+][IO3^-] /[HIO3]
0.17 = [x][x]/[0.50-x]
0.17 * 0.50 –x = x^2
Solving for x we get
0.22 M= x
So the equilibrium concentration of the H+ = x = 0.22 M
Q15) 1 atm = 29.92 inHg
26.6 inHg * 1atm / 29.92 in Hg = 0.889 atm
So the answer is 0.889 atm that is option d
Q16) half life = 5730 yr
Rate constant K = 0.693 / half life
= 0.693 / 5730 yr
= 0.000121 yr-1
Activity reduced to 0.200 % of original
Ln([A]t/[A]o)= - K*t
Ln[0.200 /100] = -0.000121 yr-1 * t
-6.2146 = -0.000121 yr-1 * t
t = 6.2146/0.000121 yr-1
t= 51400 yr
so the answer is option a = 51400 yr
Q17) process is always spontaneous when the delta H is negative that is less than 0 and delta S is positive that is greater than 0
So the answer is option C
Q18) when the vaoltaic cell reaches the equilibrium then Ecell = 0 so the answer is option C
Q19) Using the Arrhenius equation we can calculate the activation energy
Ln [K2/K1] = Ea / R [(1/T1)-(1/T2)]
Ln[0.739/0.730] = Ea / 8.314 J per mol K [(1/250)-(1/450)]
0.01225 = (Ea/ 8.314 J per mol K ) *0.00178
0.01225*8.314 J per mol K / 0.00178 = Ea
57.3 J/mol = Ea
So the answer is option C
Q20) oxidation is loss of electrons which causes the increase in oxidation number so the answer is option B
Q21) system is heated means heat is added so q= 35.0 J and it does work 15.0 J means it is negative work done
So
Delta E = q + w
= +35.0 J + (-15.0 J)
= +20.0 J
So the answer is option D
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.