B. Carbon dioxide bexause it is 96.5%
Explanation:
<h2>The number of energy levels (n) increases, so there is a greater distance between the nucleus and the outermost orbital.</h2>
Answer:
pH = 8.34
Explanation:
The equilbriums of the amphoteric HCO₃⁻ (Ion of NaHCO₃) are:
H₂CO₃ ⇄ <em>HCO₃⁻</em> + H⁺ Ka1 <em>-Here, HCO₃⁻ is acting as a base-</em>
<em>HCO₃⁻</em>⇄ CO₃²⁻ + H⁺ Ka2 <em>-Here, is acting as an acid-</em>
Where Ka1 = 4.3x10⁻⁷ and Ka2 = 4.8x10⁻¹¹. As pKa = -log Ka:
pKa1 = 6.37; pKa2 = 10.32
As the pH of amphoteric salts is:
pH = (pKa1 + pKa2) / 2
<h2>pH = 8.34</h2>
Answer: 1.27 bar
Explanation:
1 atm = 1.01325 bar
1.25 atm = Z (let Z be the unknown value)
To get the value of Z, cross multiply
Z x 1 atm = 1.25 atm x 1.01325 bar
1 atm•Z = 1.2665625 atm•bar
To get the value of Z, divide both sides by 1 atm
1 atm•Z/1 atm = 1.2665625 atm•bar/1atm
Z = 1.2665625 bar
(Round up Z to the nearest hundredth as 1.27 bar)
Thus, 1.25 atm when coverted gives 1.27 bar
