Energy, in a form of gamma rays, is <span>released when an electron and its antiparticle (positron) annihilate each other. </span>
Answer:
Ka = 3.50x10⁻⁴
Explanation:
First, we need to convert the unit of 3.60 g/L to mol/L:
The reaction dissociation of aspirin in water is:
C₉H₈O₄ + H₂O ⇄ C₉H₇O₄⁻ + H₃O⁺
0.02 - x x x
The constant of the above reaction is:
![Ka = \frac{[C_{9}H_{7}O_{4}^{-}][H_{3}O^{+}]}{[C_{9}H_{8}O_{4}]}](https://tex.z-dn.net/?f=%20Ka%20%3D%20%5Cfrac%7B%5BC_%7B9%7DH_%7B7%7DO_%7B4%7D%5E%7B-%7D%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BC_%7B9%7DH_%7B8%7DO_%7B4%7D%5D%7D%20)
To find Ka we need to find the value of x. We know that pH = 2.6 so:
![pH = -log[H_{3}O^{+}]](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20)
![x = 2.51 \cdot 10^{-3} M = [H_{3}O^{+}] = [C_{9}H_{7}O_{4}^{-}]](https://tex.z-dn.net/?f=%20x%20%3D%202.51%20%5Ccdot%2010%5E%7B-3%7D%20M%20%3D%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20%3D%20%5BC_%7B9%7DH_%7B7%7DO_%7B4%7D%5E%7B-%7D%5D%20)
Now, the concentration of C₉H₈O₄ is:
Finally, Ka is:
![Ka = \frac{[C_{9}H_{7}O_{4}^{-}][H_{3}O^{+}]}{[C_{9}H_{8}O_{4}]} = \frac{(2.51 \cdot 10^{-3})^{2}}{0.018} = 3.50 \cdot 10^{-4}](https://tex.z-dn.net/?f=%20Ka%20%3D%20%5Cfrac%7B%5BC_%7B9%7DH_%7B7%7DO_%7B4%7D%5E%7B-%7D%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BC_%7B9%7DH_%7B8%7DO_%7B4%7D%5D%7D%20%3D%20%5Cfrac%7B%282.51%20%5Ccdot%2010%5E%7B-3%7D%29%5E%7B2%7D%7D%7B0.018%7D%20%3D%203.50%20%5Ccdot%2010%5E%7B-4%7D%20)
Therefore, the Ka of aspirin is 3.50x10⁻⁴.
I hope it helps you!
C, because that is usually what waves are meant for. :)
Answer:
#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.
Explanation:
Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆
1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.
=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP
2. Calculate moles of F₂ used
=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used
3. Calculate moles of XeF₆ produced from reaction ratios …
Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆
4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number (6.02 x 10²³ molecules/mole)
=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)
= 2.75 x 10²³ molecules XeF₆.
