Answer:
hahah animals in the world
Hey there!
C₇H₉+ HNO₃ → C₇H₆(NO₂)₃+ H₂O
Our elements: C, H, N, and O.
Balance H.
Ten on the left, eight on the right. Add a coefficient of 2 in front of H₂O.
C₇H₉+ HNO₃ → C₇H₆(NO₂)₃+ 2H₂O
Balance N.
One on the left, three on the right. Add a coefficient of 3 in front of HNO₃.
C₇H₉+ 3HNO₃ → C₇H₆(NO₂)₃+ 2H₂O
This unbalanced H, let's increase the coefficient in front of H₂O from 2 to 3 to rebalance.
C₇H₉+ 3HNO₃ → C₇H₆(NO₂)₃+ 3H₂O
Balance O.
Nine on the left, nine on the right. Already balanced.
Balance C.
Seven on the left, seven on the right. Already balanced.
Our final balanced equation:
C₇H₉+ 3HNO₃ → C₇H₆(NO₂)₃+ 3H₂O
Hope this helps!
Answer:
c i believe
Explanation:
Good luck and sorry if im wrong
Answer : The pH of the solution is, 3.37
Explanation : Given,
![pK_a=3.8](https://tex.z-dn.net/?f=pK_a%3D3.8)
Concentration of
= 0.3 M
Concentration of
= 0.8 M
Now we have to calculate the pH of solution.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[HCOOK]}{[HCOOH]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BHCOOK%5D%7D%7B%5BHCOOH%5D%7D)
Now put all the given values in this expression, we get:
![pH=3.8+\log (\frac{0.3}{0.8})](https://tex.z-dn.net/?f=pH%3D3.8%2B%5Clog%20%28%5Cfrac%7B0.3%7D%7B0.8%7D%29)
![pH=3.37](https://tex.z-dn.net/?f=pH%3D3.37)
Thus, the pH of the solution is, 3.37
Answer:
Atomic radius of Strontium is 27.38pm
Explanation:
In a face-centered cubic structure, the edge, a, could be obtained using pythagoras theorem knowing the hypotenuse of the unit cell, b, is equal to 4r:
a² + a² = b² = (4r)²
2a² = 16r²
a = √8 r
As edge length of Strontium is 77.43pm:
77.43pm / √8 = r
27.38pm = r
<h3>Atomic radius of Strontium is 27.38pm</h3>