Answer:

Explanation:
Given that:-
![rate = k[NH_4^+][NO_2^-]](https://tex.z-dn.net/?f=rate%20%3D%20k%5BNH_4%5E%2B%5D%5BNO_2%5E-%5D)
k = 
![[NH_4^+]=0.301\ M](https://tex.z-dn.net/?f=%5BNH_4%5E%2B%5D%3D0.301%5C%20M)
![[NO_2^-]=0.160\ M](https://tex.z-dn.net/?f=%5BNO_2%5E-%5D%3D0.160%5C%20M)
So,

<u>
is the rate of the reaction at that temperature if [NH4+] = 0.301 M and [NO2−] = 0.160 M.</u>
Answer:
3) 2KOH + H2SO4 → K2SO4 + 2H2O
Explanation:
The balanced chemical equation for the reaction of KOH and H2SO4 is,
→ 2KOH + H2SO4 → K2SO4 + 2H2O
Hence, option (3) is the correct answer.
Answer:
Sulfur
Explanation:
<u>Valence electrons</u> are the group numbers. The group numbers are the numbers that go from left to right on the top. But since you can't have more than 8 valence electrons, the group numbers of 13-18, the valence electrons will actually be from 3-8. So 6 valence electrons would be group 16. That narrows it down a lot. The only elements in this group are oxygen, sulfur, selenium, tellurium, polonium, and livermorium.
Energy level are the period numbers. The period numbers are the numbers that go from top to bottom. So the 3rd energy level is the 3rd period. So when you look at the periodic table, under the 16 group and 3rd period, lies Sulfur.
the balanced equation for oxidation of isoborneol by naocl is written as follows
C10H18O + NaOCl ----> C10H16O + NaCl + H2O
isoborneol (C10H18O) react with NaOcl to form C10H16O NaCl and H2O