In [1]:
from xv.chemistry.physical import ReactivityManager
In [2]:
ke = ReactivityManager()
ke
Out[2]:
1455851185152@ReactivityManager
Details of elements
Minimum Grade: 6
Maximum Grade: 12
Examples
--------
ke = ReactivityManager()
ke
ke.printProblemTypes()
ke.getRandomProblem()
ke.getRandomProblem(problem_type = 0)
...
ke.printProblem()
ke.printAnswer()
ke.printSolution()
doc_style: xv_doc
Details of elements
Minimum Grade: 6
Maximum Grade: 12
Examples
--------
ke = ReactivityManager()
ke
ke.printProblemTypes()
ke.getRandomProblem()
ke.getRandomProblem(problem_type = 0)
...
ke.printProblem()
ke.printAnswer()
ke.printSolution()
doc_style: xv_doc
In [3]:
ke.printProblemTypes()
0. _problem_misc 1. _problem_reactive_1 2. _problem_reactive_2 3. _problem_activity_Na_K_Cl 4. _problem_activity_Na_ions_K_ions_electron 5. _problem_activity_Na_ions_K 6. _problem_define_oxidation_reduction 7. _problem_oxidation_or_reduction 8. _problem_define_acid 9. _problem_acid_or_base 10. _problem_activity_F_Cl_electron 11. _problem_activity_F_Cl_gas 12. _problem_electrolysis_Na_ions_K_ions 13. _problem_electrolysis_gold_plating
In [4]:
from IPython.display import HTML
n = len(ke._problemTemplates)
for j in range(0, 1):
for i in range(n):
problem_type = i
display(HTML(f"<h2>problem_type: {problem_type}/{n-1} (of loop {j})</h2>"))
ke.getRandomProblem(problem_type = problem_type, verbose = True)
display(ke.printProblem())
display(HTML(f"<h6>Answer:</h6>"))
display(ke.printAnswer())
display(HTML(f"<h6>Solution:</h6>"))
display(ke.printSolution())
pass
problem_type: 0/13 (of loop 0)
Problem Template: _problem_misc
1. Which is more reactive?
Li, Na, K
Be, Mg, Ca
2. Which is more reactive?
F, Cl, Br
O, S, Se
3. If we have Na and K atoms and one Cl atom comes in contact with them, what will happen?
4. If we have Na+ and K+ ions and one electron comes in contact with them, what will happen?
5. If we have Na+ ions and K atoms are mixed together, what will happen? Will it absorb or release energy?
6. If we have Na atoms and K+ ions are mixed together, what will happen? Will it absorb or release energy?
7. What is oxidation and reduction?
8. Is this Oxidation or reduction?
H -> H+ + e-
H+ + e- -> H
Na+ + K -> Na + K+
F + e- -> F-
F- -> F + e-
Cl- + F -> Cl + F-
H2SO4 + NaOH -> Na2SO4 + H2O
Fe + O2 -> FeO
Fe + O2 -> Fe2O3
FeO + O2 -> Fe2O3
9. What is an acid?
10. Is this acid or base?
H
H+
Cl
CL-
Na+ and K+ mixture
OH-
11. If we intorduce an electron into a mixture of chlorine and flourine atoms what is likely to happen?
12 If I give energy to a mixture of chlorine and flourine gas, which bond is likely to break first?
13. In Electrolysis, the electrolyte is a solution Na+ and K+. If current is passed what will happen?
14. If we want to electroplate a silver bar with gold coating. What kind of electrolyte and electrode?
15. Arrange in the order of strong acidty.
HCl(g)
HCl(a)
H2SO4(a)
HNO3(a)
16. What is bond energy?
https://courses.lumenlearning.com/cheminter/chapter/bond-energy/
17. What is bond disassociation/formation energy?
18. What is effect of catalist on bond disassociation/formation energy?
Answer:
Solution:
problem_type: 1/13 (of loop 0)
Problem Template: _problem_reactive_1
1. Which is more reactive?
a. Li, Na, K
b. Be, Mg, Ca
Answer:
a. Li
b. Be
b. Be
Solution:
As these options (a) have 1 valence electron and option (b) has 2 valence electron. We'll use this example:
The potential energy of the electron is zero at infinite distance from the atomic nucleus.
So taking an electron which is farthest from the atomic nucleus to infinite distance requires less energy
than taking an electron which is nearest to nucleus to infinite distance. Therefore, electron which is the farthest atomic nucleus will
be easier to release and will be most reactive.
a.
Li: 1s2 2s1
Na: 1s2 2s2 2p6 3s1
K: 1s2 2s2 2p6 3s2 3p6 4s1
As the last valence electron of K(potassium) is farthest from the atomic nucleus. K is the most reactive.
b.
Be: 1s2 2s2
Mg: 1s2 2s2 2p6 3s2
Ca: 1s2 2s2 2p6 3s2 3p6 4s2
As the last valence electrons of Ca(calcium) is farthest from the atomic nucleus. Ca is the most reactive.
The potential energy of the electron is zero at infinite distance from the atomic nucleus.
So taking an electron which is farthest from the atomic nucleus to infinite distance requires less energy
than taking an electron which is nearest to nucleus to infinite distance. Therefore, electron which is the farthest atomic nucleus will
be easier to release and will be most reactive.
a.
Li: 1s2 2s1
Na: 1s2 2s2 2p6 3s1
K: 1s2 2s2 2p6 3s2 3p6 4s1
As the last valence electron of K(potassium) is farthest from the atomic nucleus. K is the most reactive.
b.
Be: 1s2 2s2
Mg: 1s2 2s2 2p6 3s2
Ca: 1s2 2s2 2p6 3s2 3p6 4s2
As the last valence electrons of Ca(calcium) is farthest from the atomic nucleus. Ca is the most reactive.
problem_type: 2/13 (of loop 0)
Problem Template: _problem_reactive_2
2. Which is more reactive?
a. F, Cl, Br
b. O, S, Se
Answer:
a. F
b. O
b. O
Solution:
As these options (a) requires 1 electron and option (b) reqiures 2 electron. We'll use this example:
The electron wants to be in the most stable(lowest energy) state. The lowest energy state of an electron in an atom is closest possible
empty place to the atomic nucleus. Therefore, the atom which has the closest possible empty place to an electron fill up
will be the most reactive.
a.
F: 1s2 2s2 2p5
Cl: 1s2 2s2 2p6 3s2 3p5
Br: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5
As the closest possible empty place for an electron to fill up is in F(Fluorine). F is the most reactive.
b.
O: 1s2 2s2 2p4
S: 1s2 2s2 2p6 3s2 3p4
Se: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4
As the closest possible empty place for electrons to fill up is in O(Oxygen). Se is the most reactive.
The electron wants to be in the most stable(lowest energy) state. The lowest energy state of an electron in an atom is closest possible
empty place to the atomic nucleus. Therefore, the atom which has the closest possible empty place to an electron fill up
will be the most reactive.
a.
F: 1s2 2s2 2p5
Cl: 1s2 2s2 2p6 3s2 3p5
Br: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5
As the closest possible empty place for an electron to fill up is in F(Fluorine). F is the most reactive.
b.
O: 1s2 2s2 2p4
S: 1s2 2s2 2p6 3s2 3p4
Se: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4
As the closest possible empty place for electrons to fill up is in O(Oxygen). Se is the most reactive.
problem_type: 3/13 (of loop 0)
Problem Template: _problem_activity_Na_K_Cl
3. If we have Na and K atoms and one Cl atom comes in contact with them, what will happen?
Answer:
Na + K$^+$ + Cl$^+$
Solution:
K and Na each want to release an electron, and Cl wants an electron. As K is more reactive than Na, K will give an electron to Cl.
Na + K + Cl -> Na + K$^+$ + Cl$^-$
Na + K + Cl -> Na + K$^+$ + Cl$^-$
problem_type: 4/13 (of loop 0)
Problem Template: _problem_activity_Na_ions_K_ions_electron
4. If we have Na+ and K+ ions and one electron comes in contact with them, what will happen?
Answer:
Na + K$^+$
Solution:
Na$^+$ and K$^+$ each want an electron to neutralize the charge. As K$^+$ is more positive than Na$^+$, Na$^+$ will take the electron.
Na$^+$ + K$^+$ + e$^-$ -> Na + K$^+$
Na$^+$ + K$^+$ + e$^-$ -> Na + K$^+$
problem_type: 5/13 (of loop 0)
Problem Template: _problem_activity_Na_ions_K
5. If we have Na+ ions and K atoms are mixed together, what will happen? Will it absorb or release energy?
Answer:
Na + K$^+$
Solution:
K wants to release the valence electron, as it is High reactive.
Na$^+$ + K -> Na$^+$ + K$^+$ + e$^-$
As K$^+$ is more positive than Na$^+$, Na$^+$ will take the electron.
Na$^+$ + K$^+$ + e$^-$ -> Na + K$^+$
Na$^+$ + K -> Na$^+$ + K$^+$ + e$^-$
As K$^+$ is more positive than Na$^+$, Na$^+$ will take the electron.
Na$^+$ + K$^+$ + e$^-$ -> Na + K$^+$
problem_type: 6/13 (of loop 0)
Problem Template: _problem_define_oxidation_reduction
7. What is oxidation and reduction?
Answer:
Oxidation is the process of giving/releasing of electrons from the atom.
Reduction is the process of taking of electrons.
Reduction is the process of taking of electrons.
Solution:
Sodium Oxide (Na2O), Sodium is bonded with Oxygen.
Sodium is oxidised and sodium is giving/releasing electrons.
Oxidation is the process of giving/releasing of electrons from the atom.
Oxygen is reduced and oxygen is taking electrons.
Reduction is the process of taking of electrons.
Sodium is oxidised and sodium is giving/releasing electrons.
Oxidation is the process of giving/releasing of electrons from the atom.
Oxygen is reduced and oxygen is taking electrons.
Reduction is the process of taking of electrons.
problem_type: 7/13 (of loop 0)
Problem Template: _problem_oxidation_or_reduction
8. Is this Oxidation or reduction?
a. H -> H+ + e-
b. H+ + e- -> H
c. Na+ + K -> Na + K+
d. F + e- -> F-
e. F- -> F + e-
f. Cl- + F -> Cl + F-
g. H2SO4 + NaOH -> Na2SO4 + H2O
h. Fe + O2 -> FeO
i. Fe + O2 -> Fe2O3
f. FeO + O2 -> Fe2O3
Answer:
a. H is oxidised.
b. H+ is reduced.
c. Na+ is reduced. K is oxidised.
d. F is reduced.
e. F- is oxidised.
f. Cl- is oxidised. F is reduced.
g. H+ is reduced. OH- is oxidised.
h. Fe is oxidised. O2 is reduced.
i. Fe is oxidised. O2 is reduced.
f. FeO is oxidised. O2 is reduced.
b. H+ is reduced.
c. Na+ is reduced. K is oxidised.
d. F is reduced.
e. F- is oxidised.
f. Cl- is oxidised. F is reduced.
g. H+ is reduced. OH- is oxidised.
h. Fe is oxidised. O2 is reduced.
i. Fe is oxidised. O2 is reduced.
f. FeO is oxidised. O2 is reduced.
Solution:
a. H -> H+ + e-
H is oxidised.
b. H+ + e- -> H
H+ is reduced.
c. Na+ + K -> Na + K+
Na+ is reduced. K is oxidised.
d. F + e- -> F-
F is reduced.
e. F- -> F + e-
F- is oxidised.
f. Cl- + F -> Cl + F-
Cl- is oxidised. F is reduced.
g. H2SO4 + NaOH -> Na2SO4 + H2O
H+ is reduced. OH- is oxidised.
h. Fe + O2 -> FeO
Fe is oxidised. O2 is reduced.
i. Fe + O2 -> Fe2O3
Fe is oxidised. O2 is reduced.
f. FeO + O2 -> Fe2O3
FeO is oxidised. O2 is reduced.
H is oxidised.
b. H+ + e- -> H
H+ is reduced.
c. Na+ + K -> Na + K+
Na+ is reduced. K is oxidised.
d. F + e- -> F-
F is reduced.
e. F- -> F + e-
F- is oxidised.
f. Cl- + F -> Cl + F-
Cl- is oxidised. F is reduced.
g. H2SO4 + NaOH -> Na2SO4 + H2O
H+ is reduced. OH- is oxidised.
h. Fe + O2 -> FeO
Fe is oxidised. O2 is reduced.
i. Fe + O2 -> Fe2O3
Fe is oxidised. O2 is reduced.
f. FeO + O2 -> Fe2O3
FeO is oxidised. O2 is reduced.
problem_type: 8/13 (of loop 0)
Problem Template: _problem_define_acid
9. What is an acid?
Answer:
Acids is which provides H+ ions or protons or which takes electrons.
Solution:
HCl and H2SO4 are acids. In solution HCl breaks into H+ and Cl-, and H2SO4 breaks into H+ and SO4--.
Acids is which provides H+ ions or protons or which takes electrons.
Acids is which provides H+ ions or protons or which takes electrons.
problem_type: 9/13 (of loop 0)
Problem Template: _problem_acid_or_base
10. Is this acid or base?
a. H
b. H+
c. Cl
d. CL-
e. Na+ and K+ mixture
f. OH-
Answer:
a. Netural
b. Acid
c. Acid
d. Base
e. Acid
f. Base
b. Acid
c. Acid
d. Base
e. Acid
f. Base
Solution:
a. H
Netural
b. H+
Acid
c. Cl
Acid
d. CL-
Base
e. Na+ and K+ mixture
Acid
f. OH-
Base
Netural
b. H+
Acid
c. Cl
Acid
d. CL-
Base
e. Na+ and K+ mixture
Acid
f. OH-
Base
problem_type: 10/13 (of loop 0)
Problem Template: _problem_activity_F_Cl_electron
11. If we introduce an electron into a mixture of chlorine and fluorine atoms what is likely to happen?
Answer:
F$^-$ + Cl
Solution:
As F(Fluorine) is more reactive than Cl(Chlorine). The electron will be taken by the F.
F + Cl + e$^-$ -> F$^-$ + Cl
F + Cl + e$^-$ -> F$^-$ + Cl
problem_type: 11/13 (of loop 0)
Problem Template: _problem_activity_F_Cl_gas
12. If I give energy to a mixture of chlorine and flourine gas, which bond is likely to break first?
Answer:
The F molecule will break with the least energy appiled.
Solution:
F(Fluorine) is smaller than Cl(Chlorine). So the bond length between F molecule will be
smaller then bond length in Cl molecule. The repulsion between electrons in F molecule will be greater.
The F molecule will break with the least energy appiled, so F molecule bond energy will less than Cl molecule.
Bond energies:
F–F: 157 kJ/mol
Cl–Cl: 243 kJ/mol
smaller then bond length in Cl molecule. The repulsion between electrons in F molecule will be greater.
The F molecule will break with the least energy appiled, so F molecule bond energy will less than Cl molecule.
Bond energies:
F–F: 157 kJ/mol
Cl–Cl: 243 kJ/mol
problem_type: 12/13 (of loop 0)
Problem Template: _problem_electrolysis_Na_ions_K_ions
13. In Electrolysis, the electrolyte is a solution Na+ and K+. If current is passed what will happen?
Answer:
The electroplating will be done of Na(Sodium).
Solution:
In Electrolysis, there are 2 electrode and DC current passed through them. So there will be one anode(positive electrode) and
one cathode(negative electrode). Both electrodes will be submerged in the electrolyte contaning Na+ and K+ ions.
As current passes through electrodes, all electrons will be at cathode. As K$^+$ is more positive than Na$^+$, Na$^+$ will
be attracted to cathod, take the electron and form Na on the electrode.
Therefore, the electroplating will be done of Na(Sodium).
one cathode(negative electrode). Both electrodes will be submerged in the electrolyte contaning Na+ and K+ ions.
As current passes through electrodes, all electrons will be at cathode. As K$^+$ is more positive than Na$^+$, Na$^+$ will
be attracted to cathod, take the electron and form Na on the electrode.
Therefore, the electroplating will be done of Na(Sodium).
problem_type: 13/13 (of loop 0)
Problem Template: _problem_electrolysis_gold_plating
15. Arrange in the order of strong acidty.
HCl(g), HCl(a), HNO3(a), H2SO4(a)
Answer:
Acidty order: HCl(a), H2SO4(a), HNO3(a), HCl(g)
Solution:
HCl(g), HCl(a), HNO3(a), H2SO4(a)
Acidty order: HCl(a), H2SO4(a), HNO3(a), HCl(g)
Acidty order: HCl(a), H2SO4(a), HNO3(a), HCl(g)
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