Answer: a) ![[OH^-]=1.09\times 10^{-13}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.09%5Ctimes%2010%5E%7B-13%7D)
and pH = 1.04
b) ![[H^+]=1.02\times 10^{-11}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D1.02%5Ctimes%2010%5E%7B-11%7D)
and 
c) ![[H^+]=7.08\times 10^{-11}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D7.08%5Ctimes%2010%5E%7B-11%7D)
and ![[OH^-]=1.41\times 10^{-4}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.41%5Ctimes%2010%5E%7B-4%7D)
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)
a) ![[H^+]=0.090M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.090M)
![pH=-\log [0.090]=1.04](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5B0.090%5D%3D1.04)
![12.96=-log[OH^-]](https://tex.z-dn.net/?f=12.96%3D-log%5BOH%5E-%5D)
b) ![[OH^-]=0.00098M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.00098M)
![pOH=-\log [0.00098]=3.01](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5B0.00098%5D%3D3.01)
![10.99=-log[H^+]](https://tex.z-dn.net/?f=10.99%3D-log%5BH%5E%2B%5D)
c) 
![10.15=-\log [H^+]](https://tex.z-dn.net/?f=10.15%3D-%5Clog%20%5BH%5E%2B%5D)
![3.85=-log[OH^-]](https://tex.z-dn.net/?f=3.85%3D-log%5BOH%5E-%5D)
Group 1-lose 1; Group 2-lose 2; Group 13-lose 3; Group 16-gain 2; Group 17-gain 1; Group 18-gain 0
Answer:
43.75 ml
Explanation:
Given that the equation of the reaction is;
2HNO3(aq) + Ca(OH)2(aq) ---> Ca(NO3)2(aq) + 2 H20(l)
Concentration of acid CA= 0.05 M
Concentration of base CB = 0.02 M
Volume of acid VA = 35.00ml
Volume of base VB= ???
Number of moles of acid NA= 2
Number of moles of base NB=1
From
CAVA/CBVB= NA/NB
Making VB the subject of the formula;
VB= CAVANB/CBNA
VB= 0.05 × 35 × 1/ 0.02 × 2
VB=1.75 /0.04
VB= 43.75 ml
Answer:
Maybe A)
Explanation:
Can you make me the Brainliest answer ?
Answer: A. It reflects sunlight
hope this helps let me know if it's wrong
