Answer:
basic solution.
Explanation:
has a pH of 12.
it turns clear indicator pink.
it contains hydroxide ions which conduct electricity
The atomic number for Pb is 82
∴ Pb has 82 protons and 206-82 = 14 protons
The actual mass of Pb nuclei is
=(82 × mass of the proton) + (124 × mass of neutron)
=(82× 1.00728) + (124 × 1.008664) amu
= 207.6713 amu
The mass of lead which is given is 205.9744 amu
∴mass defect is
m = 207.6713 - 205.9744 = 1.6969 amu
=1.6969 × 1.66054 × 10⁻²⁷kg
=2.818 × 10⁻²⁷kg
The binding energy is E = mc²
C is the speed of light in vacuum = 2.9979 × 10⁸m/s
∴ E = 2.532 × 10×⁻¹⁰ J/mol
= 2.532 × 10⁻¹⁰ × 6.023 × 10²³ J/mol
= 1.53811 × 10¹⁴ J/mol
Both ehhevshahahbsbdvhshs
The unhybridized pz orbitals on each carbon overlap to a π bond (pi).The sigma bond framework of the ethylene molecule is produced by the overlap of hybrid orbitals or by the interaction of a hybrid orbital and a 1s hydrogen orbital.
Each carbon still has its unhybridized pz orbital, though. Sigma bond are typically the only types of single bonding between atoms. One sigma bond and two pi bonds make up triple bonds. One sigma () bond makes up a single bond, one and one pi () bond makes up a double bond, and one and two bonds make up a triple bond.
To learn more about bond, click here.
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The dissociation of formic acid is:
The acid dissociation constant of formic acid, 
is:
![k_a = \frac{[HCOO^{-}] [H^{+}]}{HCOOH}](https://tex.z-dn.net/?f=%20k_a%20%3D%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%20%20%5BH%5E%7B%2B%7D%5D%7D%7BHCOOH%7D%20%20%20%20%20)
Rearranging the equation:
![\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%7D%7B%5BHCOOH%5D%7D%20%3D%20%5Cfrac%7Bk_a%7D%7B%5BH_%2B%5D%7D%20)
pH = 2.75
![pH = -log[H^{+}]](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%5BH%5E%7B%2B%7D%5D%20)
![[H^{+}]= 10^{-2.75} = 1.78 \times 10^{-3}](https://tex.z-dn.net/?f=%20%5BH%5E%7B%2B%7D%5D%3D%2010%5E%7B-2.75%7D%20%3D%201.78%20%5Ctimes%2010%5E%7B-3%7D%20)
Substituting the values in the equation:
![\frac{[HCOO^{-}]}{[HCOOH]} = \frac{1.78\times 10^{-4}}{1.78\times 10^{-3}}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%7D%7B%5BHCOOH%5D%7D%20%3D%20%5Cfrac%7B1.78%5Ctimes%2010%5E%7B-4%7D%7D%7B1.78%5Ctimes%2010%5E%7B-3%7D%7D%20%20%20)
Hence, the ratio is 
.
