C₆H₁₂O₆, or glucose, is oxidized in the presence of oxygen to form carbon dioxide and water. The reaction equation for this is:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
Therefore, if 6 moles of oxygen are consumed, we can see from the equation that one mole of C₆H₁₂O₆ will be consumed.
Answer:
B and C
Explanation:
When we have to do a buffer solution we always have to choose the reaction that has the <u>pKa closer to the desired pH value</u>. When we find the pKa values we will obtain:
![pKa_1=-Log[6.9x10^-^3]=2.16](https://tex.z-dn.net/?f=pKa_1%3D-Log%5B6.9x10%5E-%5E3%5D%3D2.16)
![pKa_2=-Log[6.2x10^-^8]=7.20](https://tex.z-dn.net/?f=pKa_2%3D-Log%5B6.2x10%5E-%5E8%5D%3D7.20)
![pKa_3=-Log[4.8x10^-^13]=12.31](https://tex.z-dn.net/?f=pKa_3%3D-Log%5B4.8x10%5E-%5E13%5D%3D12.31)
The closer value is pKa2 with a value of 7.2. Therefore we have to use the second reaction. In which
is the <u>acid</u> and
is the <u>base</u>. Therefore the answer for the first question is B and the answer for the second question is C.
A 20 L sample of the gas contains 8.3 mol N₂.
According to <em>Avogadro’s Law,</em> if <em>p</em> and <em>T</em> are constant
<em>V</em>₂/<em>V</em>₁ = <em>n</em>₂/<em>n</em>₁
<em>n</em>₂ = <em>n</em>₁ × <em>V</em>₂/<em>V</em>₁
___________
<em>n</em>₁ = 0.5 mol; <em>V</em>₁ = 1.2 L
<em>n</em>₂ = ?; <em>V</em>₂ = 20 L
∴<em>n</em>₂ = 0.5 mol × (20 L/1.2 L) = 8.3 mol