B. When scientific understanding changes.
<u>Answer:</u> The weak bond having slightly positive region and a slightly negative region is polar covalent bond.
<u>Explanation:</u>
Covalent bond is defined as the bond which is formed by the sharing of electrons between the atoms. <u>For Example:</u> HCl,
etc..
They are of two types:
- <u>Polar covalent bond:</u> This bond is formed when difference in electronegativity between the atoms is present. When atoms of different elements combine, it results in the formation of polar covalent bond.
In this bond, a dipole is created. More electropnegative atom will have a slight negative charge and less electronegative atom will have a slight positive charge. For Example:
etc..
- <u>Non-polar covalent bond:</u> This bond is formed when there is no difference in electronegativity between the atoms. When atoms of the same element combine, it results in the formation of non-polar covalent bond. For Example:
etc..
Hence, the weak bond having slightly positive region and a slightly negative region is polar covalent bond.
Answer:
8.279
Explanation:
The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.
At the equivalence point, we have

= 25.00 x 0.200
= 5.00 m-mol
= 0.005 mol
Volume of the base that is added to reach the equivalence point is

Number of moles of 
= 0.005 mol
Volume at the equivalence point is 25 + 5 = 30.00 mL
Therefore, concentration of 
= 0.167 M
Now the ICE table :

I (M) 0.167 0 0
C (M) -x +x +x
E (M) 0.167-x x x
Now, the value of the base dissociation constant is ,



= 
Base ionization constant, ![$K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$](https://tex.z-dn.net/?f=%24K_b%20%3D%20%5Cfrac%7B%5Cleft%5BHNO_2%5Cright%5D%20%5Cleft%5BOH%5E-%20%5Cright%5D%7D%7B%5Cleft%5BNO%5E-_2%20%5Cright%5D%7D%24)


So, ![$[OH^-]=1.9054 \times 10^{-6 } \ M$](https://tex.z-dn.net/?f=%24%5BOH%5E-%5D%3D1.9054%20%5Ctimes%2010%5E%7B-6%20%7D%20%5C%20M%24)
pOH =- ![$\log[OH^-]$](https://tex.z-dn.net/?f=%24%5Clog%5BOH%5E-%5D%24)
= 
=5.72
Now, since pH + pOH = 14
pH = 14.00 - 5.72
= 8.279
Therefore the ph is 8.279 at the end of the titration.
Answer: 1-Exhale 2-inhale 3-contracted diaphragm 4-relaxed diaphragm.
Step by step explanation: hope this helped! Have a great day.