Answer:
Part A:
First, convert molarity to moles by multiplying by the volume:
0.293 M AgNO3 = (0.293 moles AgNO3)/1 L x 1.19 L = 0.349 moles AgNO3
From the calculation as shpwn in the procedure below, the equilibrium constant of the substance is 6.9 * 10^-15.
<h3>What is equilibrium constant?</h3>
The equilibrium constant for the solubility of aa solid in solution is called the solubility product Ksp. The Ksp shows the extent to which a solid is dissolved in solution.
Given that;
Fe(OH)2 ⇄Fe^2+ + 2(OH)^-
Ksp = s(2s)^2
We have s as 1.2 x 10^-5 M
So
Ksp = 4s^3
Ksp = 4( 1.2 x 10^-5 )^3
Ksp = 6.9 * 10^-15
Learn more about Ksp:brainly.com/question/27132799
#SPJ1
In the context of the 4ps of the marketing mix, high-tech media options like cell phones and the internet have had a huge impact on products.
The four Ps represent the crucial factors that must be carefully thought out and put into practice in order to properly advertise a good or service. Product, price, place, and promotion make up this list.
The marketing mix is another name for the four Ps. They cover a wide range of aspects that are taken into account when marketing a product, such as what consumers want, and how the good or service satisfies or doesn't satisfy those wants.
Also on how the good or service is perceived in society, and how it distinguishes itself from the competition, and how the business that makes it engages with its clients.
To learn more about marketing mix refer to:
brainly.com/question/14591993
#SPJ4
Answer:
Row 1
![[H^+]=1.8\times 10^{-6}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D1.8%5Ctimes%2010%5E%7B-6%7DM)
![pH=-\log[H^+]=-\log[1.8\times 10^{-6}]=5.7](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D%3D-%5Clog%5B1.8%5Ctimes%2010%5E%7B-6%7D%5D%3D5.7)
pOh=14-pH=14-5.7=8.3
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=0.5\times 10^{-8}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.5%5Ctimes%2010%5E%7B-8%7DM)
Hence, acidic
Row 2
![[OH^-]=3.6\times 10^{-10}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D3.6%5Ctimes%2010%5E%7B-10%7DM)
![pOH=-\log[OH^-]=-\log[3.6\times 10^{-10}]=9.4](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D%3D-%5Clog%5B3.6%5Ctimes%2010%5E%7B-10%7D%5D%3D9.4)
pH=14-pOH=14 - 9.4 = 4.6
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=2.6\times 10^{-5}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D2.6%5Ctimes%2010%5E%7B-5%7DM)
Hence, acidic
Row 3
pH = 8.15
![[H^+]=0.7\times 10^{-8}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.7%5Ctimes%2010%5E%7B-8%7DM)
pOH=14-pH=14 - 8.15 = 5.8
![[OH^-]=1.5\times 10^{-6}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.5%5Ctimes%2010%5E%7B-6%7DM)
Hence, basic
Row 4
pOH = 5.70
![[OH^-]=1.8\times 10^{-6}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.8%5Ctimes%2010%5E%7B-6%7DM)
pH=14-pOH=14 - 5.70 = 8.3
![[H^+]=0.5\times 10^{-8}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.5%5Ctimes%2010%5E%7B-8%7DM)
Hence, basic
Answeryou understand but id ont if your know the answer plz help me
Explanation:
