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Question
Place the following atoms in order of increasing atomic radii: Se, Sb, Br, and Te.A.Br < Se < Te < Sb
B.Se < Br < Sb < Te
C.Se < Br < Te < As
D.Sb < Te < Se < Br
E.Te < Sb < Se < Br
Answer
This answer is hidden. It contains 2 characters.
Related questions
Q:
An acid-base equilibrium system is created by dissolving 0.50 mol CH3CO2H in water to a volume of 1.0 L. What is the effect of adding 0.50 mol CH3CO2-(aq) to this solution?1/ The pH of the solution will equal 7.00 because equal concentrations of a weak acid and its conjugate base are present.2/ Some CH3CO2H(aq) will ionize, increasing the concentration of CH3CO2-(aq) and increasing the pH.3/ Some CH3CO2-(aq) will react with H3O+, increasing the concentration of CH3CO2H(aq) and reestablishing the solution equilibrium.A.1 onlyB.2 onlyC.3 onlyD.1 and 3E.1, 2, and 3
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At 25 C, all of the following ionic compounds produce a basic aqueous solution, except ____.
A.NaF
B.Na3PO4
C.NaNO2
D.NaHSO4
E.NaCH3CO2
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Hydrofluoric acid has a pKa value of 3.14. What is the value of pKb for fluoride ion?A.1.4 10-11B.7.2 10-4C.3.14D.10.86E.17.14
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Which of the following weak acids has the strongest conjugate base in an aqueous solution?A.formic acid (HCO2H), Ka = 1.8 10-4B.hypobromous (HOBr), Ka = 2.5 10-9C.hydrogen sulfite ion (HSO3-), Ka = 6.2 10-8D.nitrous acid (HNO2), Ka = 4.5 10-4E.phosphoric acid (H3PO4), Ka = 7.5 10-3
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If 0.10 M aqueous solutions are prepared of each of the following acids, which produces the solution with the lowest pH?A.benzoic acid, Ka = 6.3 10-5B.acetic acid, Ka = 1.8 10-5C.hydrocyanic acid (HCN), Ka = 4.0 10-10D.hydrogen sulfite ion (HSO3-), Ka = 6.2 10-8E.hypobromous (HOBr), Ka = 2.5 10-9
Q:
An aqueous solution with a pH of 10.60 is diluted from 1.0 L to 2.0 L. What is the pH of the diluted solution?
A.5.30
B.9.60
C.10.30
D.10.60
E.10.90
Q:
An aqueous solution with a pH of 2.00 is diluted from 1.0 L to 3.0 L. What is the pH of the diluted solution?
A.0.67
B.2.00
C.2.48
D.4.33
E.6.00
Q:
In the following reactionSO32-(aq) + H2O() HSO3-(aq) + OH-(aq)A. SO32- is and acid and HSO3- is its conjugate base.B. SO32- is a base and HSO3- is its conjugate acid.C. SO32- is a base and H2O is its conjugate acidD. SO32- is an acid and OH- is its conjugate base.E.HSO3- is an acid and OH- is its conjugate base.
Q:
Which of the following substances is never a Brnsted-Lowry acid in an aqueous solution?
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B.sodium acetate, NaCH3CO2(s)
C.ammonium nitrate, NH4NO3(s)
D.hydrogen bromide, HCl(g)
E.sodium bicarbonate, NaHCO3(s)
Q:
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Q:
Assume that the following endothermic chemical reaction is at equilibrium.
C(s) + H2O(g) H2(g) + CO(g)
Which of the following statements is/are CORRECT?
1/ Increasing the concentration of H2(g) will cause the reaction to proceed in the backward direction, increasing the equilibrium concentration of H2O(g).
2/ Decreasing the temperature will cause the reaction to proceed in the forward direction, increasing the equilibrium concentration of CO(g).
3/ Increasing the amount of C(s) will cause the reaction to proceed in the forward direction, increasing the equilibrium concentration of CO(g).
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B.2 only
C.3 only
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Q:
Given the following chemical equilibria,
N2(g) + O2(g) 2 NO(g) K1
N2(g) + 3 H2(g) 2 NH3(g) K2
H2(g) + 1/2 O2(g) H2O(g) K3
Determine the method used to calculate the equilibrium constant for the reaction below.
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) Kc
A.
B.
C.
D.
E.
Q:
The equilibrium constant (Kc) for the following reaction is 6.7 10-10 at 630 C.N2(s) + O2(g) 2 NO(g)What is the equilibrium constant for the reaction below at the same temperature?1/2 N2(g) + 1/2 O2(g) NO(g)A.4.5 10-19B.3.4 10-10C.1.3 10-9D.1.8 10-5E.2.6 10-5
Q:
The equilibrium constant at 25 C for the dissolution of silver iodide is 8.5 10-17.AgI(s) Ag+(aq) + I-(aq)If an excess quantity of AgI(s) is added to water and allowed to equilibrate, what is the equilibrium concentration of I-?A.7.2 10-33 MB.4.3 10-17 MC.8.5 10-17 MD.6.5 10-9 ME.9.2 10-9 M
Q:
At a given temperature, 0.0664 mol N2O4(g) is placed in a 1.00 L flask. After reaching equilibrium, the concentration of NO2(g) is 6.1 10-3 M. What is Kc for the reaction below?N2O4(g) 2 NO2(g)A.3.7 10-5B.1.4 10-4C.5.9 10-4D.9.6 10-2E.1.8 103
Q:
If the reaction quotient, Q, is equal to K in a gas phase reaction, then
A.the chemical system has reached equilibrium.
B.the temperature must be increased for the reaction to proceed in the forward direction.
C.the reaction will proceed in the forward direction until equilibrium is established.
D.the reaction will proceed in the backward direction until equilibrium is established.
E.the reaction will proceed in the direction that increases the number of gas phase particles.
Q:
Write a balanced chemical equation which corresponds to the following equilibrium constant expression.A.HNO2(aq) + H2O() NO2-(aq) + H3O+(aq)B.NO2-(aq) + H3O+(aq) HNO2(aq) + H2O()C.NO2-(aq) + H3O+(aq) HNO2(aq)D.H+(aq) + OH-(aq) H2O()E.HNO2(aq) NO2-(aq) + H3O+(aq)
Q:
For the reaction 2A + B C, the rate law is
.
Which of the factor(s) will affect the value of the rate constant for this reaction?
1/ decreasing the temperature
2/ adding a catalyst
3/ decreasing the concentration of reactant A
A.1 only
B.2 only
C.3 only
D.1 and 2
E.2 and 3
Q:
For the reaction A B, the rate law is
.
What are the units of the rate constant where time is measured in seconds?
A.
B.
C.
D.
E.
Q:
What is the overall order of the reaction below
NO(g) + O3(g) NO2(g) + O2(g)
if it proceeds via the following rate expression? A.zero-order
B.first-order
C.second-order
D.third-order
E.fourth-order
Q:
Which statement concerning relative rates of reaction is correct for the chemical equation given below?
2 CH3OH(g) + 3 O2(g) 2 CO2(g) + 4 H2O(g)
A.The rate of disappearance of CH3OH is equal to the rate of disapperance of O2.
B.The rate of disappearance of CH3OH is two times the rate of appearance of H2O.
C.The rate of disappearance of CH3OH is half the rate of appearance of CO2.
D.The rate of appearance of H2O is two times the rate of appearance of CO2.
E.The rate of appearance of H2O is four times the rate of disappearance of CH3OH.
Q:
Nitrogen dioxide reacts with carbon monoxide to produce nitrogen monoxide and carbon dioxide.NO2(g) + CO(g) NO(g) + CO2(g)A proposed mechanism for this reaction is2 NO2(g) NO3(g) + NO(g) (fast, equilibrium)NO3(g) + CO(g) NO2(g) + CO2(g) (slow)What is a rate law that is consistent with the proposed mechanism?A.rate = k[NO2]2[CO] [NO]-1B.rate = k[NO2]2[CO]C.rate = k[NO2][CO]D.rate = k[NO3][CO]E.rate = k[NO2]2
Q:
The mechanism of a chemical reaction is given below.(CH3)3CCl (CH3)3C+ + Cl- (slow)(CH3)3C+ + OH- (CH3)3COH (fast)Which of the following statements concerning the reaction is/are CORRECT?1/ The overall balanced reaction is: (CH3)3CCl + OH- (CH3)3COH + Cl-2/ Hydroxide ion is a reaction intermediate.3/ The following rate law is consistent with the mechanism: rate = k[(CH3)3CCl]OH-].A.1 onlyB.2 onlyC.3 onlyD.1 and 3E.1, 2, and 3
Q:
The rate constant at 366 K for a first-order reaction is 7.7 10-3 s-1 and the activation energy is 15.9 kJ/mol. What is the value of the frequency factor, A, in the Arrhenius equation? (R = 8.314 J/Kmol)A.0.0047 s-1B.0.70 s-1C.0.93 s-1D.1.1 s-1E.1.4 s-1
Q:
Calculate the activation energy, Ea, forN2O5(g) 2 NO2(g) + 1/2 O2(g)given k (at 45.0 C) = 5.79 10-4 s-1 and k (at 60.0 C) = 3.83 10-3 s-1. (R = 8.314 J/Kmol)A.0.256 kJ/molB.2.83 kJ/molC.31.1 kJ/molD.111 kJ/molE.389 kJ/mol
Q:
What is the half-life of a first-order reaction if it takes 4.4 10-2 seconds for the concentration to decrease from 0.50 M to 0.20 M?A.2.5 10-2 sB.3.3 10-2 sC.1.6 sD.21 sE.27 s
Q:
Hydrogen peroxide decomposes into water and oxygen in a first-order process.
H2O2(aq) H2O() + 1/2 O2(g)
At 20.0 C, the half-life for the reaction is 3.92 104 seconds. If the initial concentration of hydrogen peroxide is 0.52 M, what is the concentration after 7.00 days?
A.1.2 10-5 M
B.0.034 M
C.0.074 M
D.0.22 M
E.0.52 M
Q:
A student analyzed a first-order reaction and obtained the graph below. Unfortunately, the student forgot to label the axes. What are the correct labels for the x and y axes? A.x axis = time, y axis = ln[A]
B.x axis = ln[time], y axis = [A]
C.x axis = ln[time], y axis = [A]
D.x axis = time, y axis = 1/[A]
E.x axis = 1/time, y axis = 1/[A]