Answer:
1st paragraph:
1.compound 2.physical 3.elements 4.water 5. gas 6.white 7. season 8.ratio 9.formula 10.elements 11.atoms 12.subscript 13.one
2nd paragraph
1.stable 2.many/reactive 3.eight 4.helium 5.seven 6.outer 7.one 8.level 9. compounds 10.reactive 11. seven 12. Eight 13.lose 14.gain 15. Share 16.compounds 17. atoms
Explanation:
I just did the 1st page. Gimme a min and I'll do the second.
Answer : The oxidizing element is N and reducing element is O.
is act as an oxidizing agent as well as reducing agent.
Explanation :
An Oxidizing agent is the agent which has ability to oxidize other or a higher in oxidation number.
Reducing agent is the agent which has ability to reduce other or lower in oxidation number.
The given reaction is :

act as an oxidizing agent.
The oxidation number of N in
is calculated as:
(+1)+(x)+3(-2) = 0
x = +5
And the oxidation number of N in
is calculated as:
(+1)+(x)+2(-2) = 0
x = +3
From the oxidation number method, we conclude that the oxidation number reduced this means
itself get reduced to
and it can act as an oxidizing agent.
act as a reducing agent.

The oxidation number of O in
is calculated as:
(+1)+(+5)+3(x) = 0
x = -2
The oxidation number of O in
is Zero (o).
Now, we conclude that the oxidation number increases this means
itself get oxidized to
and it can act as reducing agent.
1. Convert gallons to mL. 1 gal = 3785.4117840007 mL, multiply that by 29 and get 109776.94173602 mL.
2. Since there is one gram per every mL, there are 109776.94173602 g of water in the fish tank.
3. Convert g to pounds. 1 g = 0.0022 pounds. Multiply 109776.94173602 by 0.0022 and end up with about 241.5 pounds of water.
Answer: 0.0069L
Explanation:
2H2O(l) ---->O2(g) + 4H+(aq) + 4e-
no of moles= it/eF
NO of moles of O2 produced = (Current in Ampere x Time in second)/ (Faraday constant x Number of electrons required)
Moles of O2 produced = (0.02x (60 x 60X1.5 s)/(96485 x 4)
= 0.0002798 moles= 2.798x 10 ^-4moles
Using ideal gas equation,
P V = n R T
Where, P is the pressure,
V is the volume,
n is the number of moles,
R is the gas constant, and T is the temperature
We have, 1 bar = 0.986923 atm
Substituting the values,
V = nRT/P = (2.798 x 10-4moles x 0.08205 L atm mol K x 298 K)/ 0.986923 atm = 0.0069L
Volume of O2 produced = 0.0069L