Explanation:
It is known that for 
, ppm present in 1 
are as follows.
1 
= 0.494 ppm
So, 150 
= 
= 0.15
Therefore, calculate the equivalent concentration in ppm as follows.
= 0.074 ppm
Thus, we can conclude that the equivalent concentration in ppm at STP is 0.074 ppm.
The pH of pure water has been best described as neutral pH with equal hydronium and hydroxide ions. Thus, option A is correct.
pH has been described as the measurement of hydrogen ions in a solution. The pH has been measured on a scale of 1-14. pH 7 has been the neutral pH.
The higher hydronium ion concentration tends to move the pH from 7 towards 1. The higher hydroxide ion concentration tends to move the pH above 7.
The neutral pH has been neither acidic nor basic with the equal constituents of hydronium and hydroxide ion in the solution.
Thus, the pH of pure water has been 7. It has neutral pH with equal hydronium and hydroxide ions. Thus, option A is correct.
For more information about the pH of the solution, refer to the link:
brainly.com/question/4975103
Answer: 1.96x10^24 atoms
Explanation:
3.25*6.02214076*10^23 atoms = 1.96x10^24
a. pH=2.07
b. pH=3
c. pH=8
<h3>Further explanation</h3>
pH=-log [H⁺]
a) 0.1 M HF Ka = 7.2 x 10⁻⁴
HF= weak acid
![\tt [H^+]=\sqrt{Ka.M}\\\\(H^+]=\sqrt{7.2.10^{-4}\times 0.1}\\\\(H^+]=8.5\times 10^{-3}\\\\pH=3-log~8.5=2.07](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7BKa.M%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D%5Csqrt%7B7.2.10%5E%7B-4%7D%5Ctimes%200.1%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D8.5%5Ctimes%2010%5E%7B-3%7D%5C%5C%5C%5CpH%3D3-log~8.5%3D2.07)
b) 1 x 10⁻³ M HNO₃
HNO₃ = strong acid
![\tt pH=-log[1\times 10^{-3}]=3](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B1%5Ctimes%2010%5E%7B-3%7D%5D%3D3)
c) 1 x 10⁻⁸ M HCl
![\tt pH=-log[1\times 10^{-8}]=8](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B1%5Ctimes%2010%5E%7B-8%7D%5D%3D8)
Answer:
The oxidation number of the metal decreases
2 Al + Fe₂O₃ → Al₂O₃ + 2 FeO
The metal element iron, is reduced from Fe⁺³ in Fe₂O₃ to Fe⁺² in FeO
Explanation:
When an element gains electron, the element becomes reduced, hence when a metal is reduced, the metal gains electrons, which reduces the oxidation number of the metal
An example of a metal being reduced is;
2 Al + Fe₂O₃ → Al₂O₃ + 2 FeO
In the above reaction, the iron (III) oxide is reduced to iron (II) oxide by aluminium metal.
